Vegard Aune

Folco Casadei

Georgios Valsamos

Tore Borvik

Formulation and Implementation of the VPJC Material Model in

EUROPLEXUS

2016

EUR 27982 EN

Formulation and Implementation of the VPJC Material Model in

EUROPLEXUS

This publication is a Technical report by the Joint Research Centre, the European Commission’s in-house science service. It aims to provide evidence-based scientific support to the European policy-making process. The scientific output expressed does not imply a policy position of the European Commission. Neither the European Commission nor any person acting on behalf of the Commission is responsible for the use which might be made of this publication. JRC Science Hub https://ec.europa.eu/jrc JRC102336 EUR 27982 EN ISBN 978-92-79-59746-6 ISSN 1831-9424 doi:10.2788/609529 © European Union, 2016 Reproduction is authorised provided the source is acknowledged. Printed in Italy All images © European Union 2016 How to cite: Authors; title; EUR; doi

Formulation and Implementation of the VPJC Material Model inEUROPLEXUS

Vegard Aunea,b,∗, Folco Casadeic, Georgios Valsamosc, Tore Børvika,b

aCentre for Advanced Structural Analysis (CASA), NTNU, Norwegian University of Science and Technology, NO-7491Trondheim, Norway

bStructural Impact Laboratory (SIMLab), Department of Structural Engineering, NTNU, NO-7491 Trondheim, NorwaycEuropean Commission, Joint Research Centre (JRC), Institute for the Protection and Security of the Citizen (IPSC),

European Laboratory for Structural Assessment (ELSA), 21027 Ispra, Italy

Abstract

A constitutive model of elastic-thermoviscoplasticity and ductile failure is implemented in EUROPLEXUS(EPX) using a fully vectorized version of the forward Euler integration algorithm and a special case of thebackward Euler algorithm. The material model is based on the constitutive model originally proposed byJohnson and Cook, and the ductile failure criterion proposed by Cockcroft and Latham. Constitutive equa-tions are formulated in terms of linear elasticity, von Mises yield criterion, associated flow rule, nonlinearisotropic strain hardening, strain-rate hardening, temperature softening due to adiabatic heating, and duc-tile failure. The performance of the implementation is verified and validated in EPX. First, a verificationis performed using single element simulations in uniaxial tension and simple shear. Then, the model isvalidated against experimental data from quasi-static tension tests and airblast experiments. The model isapplicable for one-, two- and three-dimensional stress analysis, and coupled with element deletion optionsavailable in EPX.

Keywords:Elastic-thermoviscoplasticity, Isotropic hardening, von Mises yield criterion, Adiabatic thermal softening,Cutting plane return map algorithm, Radial return map algorithm, VPJC, EUROPLEXUS

1. Introduction

In metal plasticity, extreme loading conditions in fast transient dynamics often involve large strains, highstrain rates, temperature softening and ductile failure. A widely used design tool for this class of problemsis the non-linear finite element (FE) method. In the FE method, the formulation and numerical solutionof non-linear problems in continuum mechanics relies on the weak form of the momentum balance equation(also known as the principal of virtual work [1]). The integral form of the principal of virtual work is wellsuited for direct application to the FE method, and by spatial discretization the solution is integrated intime. In the case of fast transient dynamics, it is often used an explicit time integration based on thecentral difference method. Iterations for FE analyses involving non-linear material behaviour can in generalbe divided into two levels, hereby denoted the global and local level. The global level involves the explicitestablishment of global equilibrium between internal stresses and external loads, and the local level updatesthe corresponding stress state in each FE integration point (for a given strain increment) in terms of thegoverning constitutive equations. Constitutive equations (also known as material models) are mathematicaldescriptions of material behaviour which gives the stress as a function of the deformation history of the body

∗Corresponding authorEmail address: vegard.aune@ntnu.no (Vegard Aune )

Technical report - Formulation and implementation of the VPJC material model June 24, 2016

[1]. This implies that the local integration of the constitutive equations controls the accuracy and stabilityof the global equilibrium, and that the overall solution in the FE model is considerably influenced by theaccuracy, robustness and effectiveness of the integration algorithm.

A commonly used scheme for the local stress integration is the predictor-corrector method (often calledreturn mapping). The basis for this return mapping is two successive steps, i.e., the predictor step and thecorrector step. The predictor step is used to estimate a trial stress state, while the corrector step applies aflow rule by using a return mapping scheme to ensure the consistency condition. That is, ensuring that thestress state is on the yield surface. The present study uses the explicit cutting plane method proposed byOrtiz and Simo [2] and the implicit closest point projection method proposed by Simo and Taylor [3]. Duringthese approaches an elastic predictor uses the total strain increment and the linear elastic relation to obtaina trial stress state. Then, if the elastic trial stress is outside the plastic domain, the elastically predictedstresses are corrected to return to the updated yield surface by iterations of the plastic corrector. The basisfor the cutting plane method is to utilize the known stress state at the last converged time increment todetermine the normal to the yield surface, while the implicit closest point projection method utilizes thecylindrical shape of the yield surface and the deviatoric stress space to scale the trial state to fit a suitablyupdated yield surface by iterations of the plastic corrector [4, 5].

This work considers a well-known constitutive relation proposed by Johnson and Cook [6] and von Misesplasticity is used to update the stresses. A model including elastic-thermoviscoplasticity and ductile failureis implemented as a new material called VPJC in EUROPLEXUS (EPX) [7]. The implementation uses afully vectorized version of the forward Euler integration algorithm and a special case of the backward Euleralgorithm. Large deformations are accounted for by using a hypoelastic-plastic formulation in the constitu-tive equations and co-rotational formulations in EPX (i.e., the Jaumann stress rates [7]). This implies thatthe constitutive equations are defined in a co-rotational coordinate system in which the basis system rotateswith the material. The subroutine interface has been implemented using the Cauchy stress components, andall stress and strain quantities are defined in terms of the rate of deformation tensor. The model is applicablefor one- (1D), two- (2D) and three-dimensional (3D) stress analyses (i.e., for bar, shell, solid, axisymmetricand plane strain elements). The model is coupled with the element deletion options available in EPX byintroducing a state variable controlling the element erosion. That is, as the failure variable reaches its criticalvalue DC the element is removed. The model is finally verified and validated through numerical simulationsin EPX. This is first done by a verification using single element simulations in uniaxial tension and simpleshear. Then, the model is validated against experimental data from quasi-static tension tests and airblastexperiments.

2. Constitutive Equations

There are in general two categories within the plasticity theory, i.e. the mathematical theories and thephysical theories [8]. Mathematical theories are purely phenomenological and aim to represent experimentalobservations through mathematical relations. Most of the common material models today are phenomeno-logical. Physical theories try to describe the underlying mechanisms, and quantify the plastic deformation,at a microscopical level. A well-known and acknowledged physical theory is the crystal plasticity theory.This section considers the phenomenological-based formulation of the constitutive equations for the compu-tational model used in this study, starting with the basic theory for infinitesimal strains and then expandingthe theory to also include large deformations. The proposed computational model includes linear elasticity,large strains, high strain rates, temperature softening and ductile failure. Assuming negligible elastic strainscompared to the plastic strains, an hypoelastic formulation may be adopted through an additive decom-position of the rate of deformation tensor. Since the main objective is the numerical implementation ofthermo-viscoplasticity, the constitutive equations are restricted to linear isotropic elastic behavior. Thermo-dynamic coupling is not included, however, the temperature evolution is assumed to be adiabatic. Ductilefailure is considered by using a simple one-parameter failure criterion suggested by Cockcroft and Latham

2

[9] which is based on plastic work per unit volume.

From a computational point of view, the plastic response is an incremental process which implies that theconstitutive equations need to be formulated on rate form. Starting by assuming infinitesimal deformations,the strain rate tensor ε may be decomposed by an additive split

ε = εe + εp (1)

where εe is the elastic and εp is the plastic strain rate tensor, respectively. Using the generalized Hooke’slaw, the rate of the stress tensor is defined as

σ = C : εe (2)

where C is the fourth order elasticity tensor given by

C = λelI⊗ I + 2µelI (3)

where I is the second order identity tensor, and λel and µel are the Lame constants given as

λel =νE

(1 + ν)(1− 2ν)and µel =

E

2(1 + ν)(4)

The elasticity tensor C is assumed to be constant, E is the Young’s modulus and ν the Poisson’s ratio.

In general, assuming only isotropic hardening, the yield criterion is defined as

f(σ, R) = ϕ(σ)− σy(σ0, R) = σeq − (σ0 +R) (5)

where ϕ(σ) = σeq is the equivalent stress, σ0 is the initial yield stress and R = R(p) is the isotropic hardeningvariable depending on the equivalent (or accumulated) plastic strain p. The evolution of the yield surface istherefore governed by the elastic domain, described by the flow stress σy. The function f is assumed to be acontinuous function of the stress tensor σ, and results in negative values when the material is in the elasticdomain. In the case of rate-independent plasticity and isotropic hardening, it is assumed that f(σ) > 0 isinadmissible and, consequently, that plastic deformation occurs when f(σ) = 0.

Furthermore, ϕ(σ) is assumed to be an isotropic and positive homogeneous function of order one of thestress tensor. This implies that

ϕ(aσ) = aϕ(σ) (6)

where a could be any positive scalar. Then, Euler’s theorem for homogeneous functions gives [10]

σ :∂ϕ

∂σ= ϕ (7)

The plastic deformation process is governed by an important physical restriction, i.e., that work has to bedone to the body all the time for the deformation to continue. This restriction is expressed as

Dp = σεp ≥ 0 (8)

where Dp expresses the plastic dissipation per unit volume, which is assumed to be dissipated as heat.

The restriction in Eq.(8) is fulfilled by defining the plastic strain rate tensor εp in a way that ensuresnon-negative dissipation. This is done by the plastic flow rule, which in the general case is expressed as

εp = λ∂g

∂σ(9)

3

where g = g(σ) ≥ 0 is the plastic potential function and λ ≥ 0 is a scalar denoted the plastic multiplier.

Combining Eqs.(8) and (9) result in the following restriction on the plastic potential function

σ :∂g

∂σ≥ 0 (10)

where it is assumed that g is a positive homogeneous function of order one. Then, by using the Euler’stheorem for homogeneous functions, it is seen that this assumption fulfills the dissipation inequality inEq.(8) as

Dp = σεp = λσ :∂g

∂σ= λg ≥ 0 (11)

By further assuming that the plastic potential function g is defined by the yield function f , Eq.(9) is expressedas

εp = λ∂f

∂σ(12)

and it is seen that the plastic potential function is associated with the yield function f . Eq.(12) is thereforecalled the associated flow rule, and ensures non-negative dissipation through Eqs.(5), (7) and (8) as

Dp = σεp = λσ :∂f

∂σ= λσ :

∂ϕ

∂σ= λϕ = λσeq ≥ 0 (13)

This implies, that in the case of the associated flow rule, the equivalent plastic strain rate equals the plasticparameter

p = λ (14)

where the equivalent plastic strain p is defined as the energy conjugate variable to the equivalent stress σeq.The plastic dissipation could then be expressed in terms of the equivalent stress and the plastic strain rateas

Dp = σεp = σeqp (15)

Hence, assuming the associated flow rule implies that the incremental plastic strain vector dεp = εpdt isparallel to the gradient of the yield surface ∇f at σ, and therefore directed along the outward normal n ofthe yield surface (see Figure 1). Moreover, the magnitude of εp is given by the plastic multiplier λ.

An important remark to this argumentation is that the shape of the yield surface also determines the direc-tion of the plastic flow, as well as the stress state at which yielding initiates.

The associated flow rule may also be derived from a stronger assumption than the dissipation inequality inEq.(8), i.e., the principle of maximum plastic dissipation expressed as

(σ − σ∗) εp ≥ 0 (16)

where σ is the stress state at which the plastic dissipation occurs, and σ∗ is an arbitrary stress state insideor on the yield surface f(σ∗) ≤ 0. As for the dissipation inequality this principle states the normality ofplastic flow, and in addition it is seen that the yield surface must be convex [11]. Thus, the use of an associ-ated flow rule ensures that the plastic strain increments dεp are vectors perpendicular to the potential surface.

To distinguish between elastic loading/unloading and plastic loading, the Kuhn-Tucker conditions are usedto determine when plastic flow may occur. These conditions determines the evolution of the yield surfaceand are given as [11]

4

σ

σ2

σ1

f(σ) = 0

dεp = dλ ∂f∂σ = dλn

∇f = ∂f∂σe

Figure 1: Geometric representation of the associated flow rule for an elliptic yield surface in plane stressspace. The orthogonal unit vectors e are pointing in the coordinate directions.

f ≤ 0, λ ≥ 0, λf = 0 (17)

where the first of these conditions indicates that the stress states must lie on or within the yield surface, thesecond indicates that the plastic multiplier is non-negative, and the last condition ensures that the stresseslie on the yield surface during plastic loading. From the Kuhn-Tucker conditions, the consistency conditioncan be stated as [10]

λf = 0 (18)

The consistency condition ensures that the stress must be located on the yield surface during continuousplastic flow, and is, in the theory of rate-independent plasticity, used to determined the plastic multiplier λ.

To account for viscoplasticity and strain-rate sensitivity, the theory of plasticity has to be extended. Themain difference is that in the viscoplastic domain the stress is allowed to move outside the yield surface, byassuming that only elastic strains are developed while the stress is inside or on the yield surface (f ≤ 0) [11].This means that plastic strains are assumed to evolve only when the stresses move outside the yield surface(Figure 2b). By using an additive constitutive relation, this implies that the yield function f is equal to theviscous stress σv(p), i.e.,

f(σ) = σv(p) > 0 ⇐⇒ fv(σ) = ϕ(σ)− σy(σ0, R)− σv(p) = 0 (19)

From Eq.(19) it is seen that the viscous stress controls the strain-rate sensitivity of the material (Figure2). Although the additive formulation is illustrative for the viscoplastic material behavior, a multiplicativeformulation may be a preferred choice in some applications. In general, the multiplicative formulation reads

fv(σ) = ϕ(σ)− σy(σ0, R)υp(p) = 0 ⇐⇒ f(σ) = σv(p, p) = (υp − 1)σy > 0 (20)

where υp(p) is determined by the constitutive relation. It should be noted that in both Eq.(19) and Eq.(20)the viscoplastic constraint fv(σ) = 0 is introduced, which allows for an expansion of the elastic domain dueto the rate sensitivity of the material. The rate sensitivity of the material is controlled by the constitutiverelation for the plastic multiplier, generally expressed as

p = λ =

0 for f ≤ 0

φ(σ, R) for f > 0(21)

5

where φ(σ, R) depends on the state of the material. From Eq.(21) it is seen that the consistency condition inEq.(18) no longer holds in the case of a viscoplastic formulation, and that the plastic flow depends explicitlyon time during the evolution of λ.

p

R(p)σ0

σv(p, p)

σeq

(a) Equivalent stress vs equivalent plastic strain

σ3

σ2σ1

π-plane

σ

f > 0f = 0

(b) The deviatoric plane (π-plane)

Figure 2: Viscoplastic effects due to strain-rate sensitivity.

To account for large displacement and large strain, the constitutive model is defined in a co-rotationalcoordinate system in which the basis system rotates with the material [7]. Typically, there are two differentapproaches of dealing with finite deformations, i.e., the hypoelastic-plastic formulations and hyperelastic-plastic formulations. In contrast to the hyperelastic formulation of finite plasticity models, hypoelastictheories do not rely on the assumption of the existence of a strain energy potential to model the reversiblebehaviour. Moreover, in the particular case of metal plasticity, the elastic strains are negligible compared tothe plastic strains and it is reasonable to adopt an hypoelastic formulation for large deformations. This isanalogous to the additive decomposition of the strain-rate tensor ε in Eq. (1) in the theory for infinitesimaldeformations, and the rate-of-deformation tensor D is decomposed additively as [1]

D = De + Dp (22)

where De and Dp are the elastic and plastic part, respectively. The rate-of-deformation tensor D is alsocalled velocity strain and is a rate measure of the deformation and stretching of the elements.

The starting point for hypoelastic models is the formulation of the constitutive equations in terms of objectivestress rates [5], in which the hypoelastic relation in terms of the objective stress rate is expressed as

σ∇ = C : De = λel (trDe) I + 2µelD

e (23)

where σ is the Cauchy stress tensor, λel and µel are the Lame constants given in Eq.(4), and σ∇ is anobjective (or frame-invariant) stress rate given as

σ∇ = σ −Aσ + σA (24)

where σ is the time derivative of the Cauchy stress tensor σ and A is an appropriate angular velocity tensor.By combining Eqs.(23) and (24) the objective stress rate is obtained as

σ = Aσ − σA+ λel (trDe) I + 2µelD

e (25)

The choice of an appropriate angular velocity tensor A is typically done by the respective FE codes, whereEPX [7] uses the Jaumann rate of the Cauchy stress σ∇J given as

6

σ∇J = σ −Wσ + σW , A = W =1

2

(L−LT

)(26)

where W is the spin tensor indicating the rate of rotation of an element and L is the velocity gradient (seee.g. [1] for more details). Another frequently used objective stress is obtained by using the Green-Naghdistress rate σ∇G [1] expressed as

σ∇G = σ − RRTσ + σRRT = σ −Ωσ + σΩ, A = Ω = RRT = −RRT (27)

where R is the rotation tensor representing the rigid body motion in the polar decomposition F = RU (seeFigure 3) and Ω is the skew-symmetric rate of the rotation tensor.

x1

x2

R

x1

x2

x2

U

x1

x2

x1

F

Figure 3: Polar decomposition of a typical homogeneous deformation. The Cartesion coordinate systemwith coordinate axes (x1, x2) refers to the global reference frame, while (x1, x2) represents the local (orco-rotated) coordinate system.

From Eqs. (26) and (27) it is seen that the Green-Naghdi rate differs from the Jaumann rate only in usinga different measure of the rotation of the material. Despite the difference in Eqs. (26) and (27), it is worthnoting that when using polar decomposition F = RU the spin tensor W can be expressed as [12]

W = RRT − 1

2R(UU−1 + U−T U

)RT (28)

where U is the symmetric right stretch tensor and it is used that F = LF . Thus, for pure rigid motion(U = 0) the Green-Naghdi and Jaumann stress rates give the same result since

W = RRT = Ω (29)

7

whereas in for instance shear one should be aware of the different formulation and how this may influencethe results of the FE simulation [1].

As concerns the plastic deformation rate tensor Dp this is defined by means of the associated flow rule as

Dp = λ∂f

∂σ(30)

2.1. The von Mises yield criterion

The von Mises yield criterion is an isotropic and pressure-insensitive criterion commonly used for metals andalloys [11]. The pressure insensitivity in metals and alloys is due to the fact that the plastic deformationto a large extent takes place by plastic slip, which is a shear driven deformation mode [10]. Geometrically,the yield criterion f(σ) = 0 defines a surface in the stress space which is defined by the components of thestress tensor σ (i.e., the vector space in Figure 4a). The elastic domain is given by the stress space insidethis yield surface, while the plastic domain is the surface itself.

e = 1√3

[1 1 1]σ3

σ2

σ1

π-plane

(Deviatoric plane)

σ1 + σ2 + σ3 = 0

Rvr

e· r

P

yield curve

yield surface

Hydrostatic axisσ1 = σ2 = σ3

(a) Principal stress space

Rv

σ3

σ2σ1

√23σy

π-plane

(b) The deviatoric plane (π-plane)

Figure 4: The von Mises yield criterion.

From Figure 4a it is seen that the von Mises yield surface is a circular cylinder of radius Rv =√

23σy.

The stress state on the yield surface could then be represented in the principal stress space as the vectorr = [σ1 σ2 σ3], which has to satisfy the following equation [12]

r · r− (e · r) (e · r) = R2v (31)

By splitting the stress in deviatoric and hydrostatic parts

σ′

= σ − σHI and σH =1

3tr(σ) =

σ1 + σ2 + σ3

3(32)

where σ′

is the deviatoric stress and σH is the hydrostatic stress, the von Mises yield criterion could beexpressed as

R2v = σ : σ −

(1√3

tr(σ)

)2

= (σ′

ij + σHδij)(σ′

ij + σHδij)− 3σ2H = σ

′

ijσ′

ij = σ′: σ′

=2

3σ2y (33)

Eq.(33) implies that isotropic (or hydrostatic) states of stress do not result in yielding. Thus, the von Misesyield criterion is purely deviatoric and additional hydrostatic stress only translates the stress state along thehydrostatic axis in Figure 4a. Furthermore, the result in Eq.(33) could be presented in terms of the yieldfunction f from Eq.(5), i.e.,

8

f(σ) = σeq − σy =

√3

2σ′ : σ′ − σy =

√3J2 − σy (34)

where the von Mises equivalent stress σeq, in terms of the stress components, reads

σeq =√σ2

11 + σ222 + σ2

33 − σ11σ22 − σ22σ33 − σ33σ11 + 3(σ212 + σ2

23 + σ231) (35)

and the associated flow rule in Eq.(30) is then given as

Dp = λ∂f

∂σ= λ

3

2

σ′

σeq(36)

The rate of the accumulated plastic strain p, in the particular case of associated flow and von Mises isotropicbehaviour, is defined as

p =

√2

3Dp : Dp (37)

2.2. The modified Johnson-Cook constitutive model

As discussed earlier, engineering structures subjected to fast transient dynamics may experience large strains,high strain rates and temperature softening. A widely used constitutive relation allowing for these factors isthe constitutive model of Johnson and Cook [6], which is empirical and dates back to 1983. The constitutiverelation has a multiplicative formulation and is given by

σy = (A+Bpn) (1 + Cln(p∗)) (1− T ∗m) (38)

where A, B, C, n and m are material constants determined from material tests, and

p∗ =p

p0and T ∗ =

T − TrTm − Tr

(39)

represents the dimensionless strain rate p∗ and the non-dimensional homologous temperature T ∗, respec-tively. The parameter p0 is a user-defined reference strain rate, while Tr and Tm are defined as the roomand melting temperature, respectively. The first term in Eq.(38) represents the strain hardening, the sec-ond term describes the influence of the strain rate, and the last term represents the temperature softeningbehaviour. The second and third term shift the hardening curve up or down, depending on the strain rateand the temperature conditions. It is seen from Eq.(38) that the flow stress σy increases with increasingplastic strain rate (which is a necessary characteristic in viscoplasticity), while it decreases with increasingtemperature. Thus, an increase in the plastic strain rate expands the yield surface whereas an increase intemperature contracts the yield surface.

To avoid non-physical softening effects when p∗ < 1, the strain-rate-sensitivity term in the Johnson-Cookmodel can be adjusted [13]. This leads to the modified Johnson-Cook model, given by [14]

σy = (A+Bpn) (1 + p∗)C (1− T ∗m) (40)

This is illustrated by a comparison of Eqs.(38) and (40) in Figure 5, where the ordinate axis refers to thenormalized stress σy/ [(A+Bpn) (1− T ∗m)]. Note that, since the formulations in Eqs.(38) and (40) areslightly different, the calibration of the viscoplastic material parameter C may be somewhat different in thetwo versions of the Johnson-Cook model.

The first term of the Johnson-Cook model could lead to numerical instabilities when n < 1 due to the initialvalue of p, i.e., p equal zero before any plasticity occur. This results in an infinite value of the hardeningmodulus HR = Bnpn−1 when plasticity occurs. This could be avoided by defining the initial plastic strain

9

10−5

10−3

100

103

105

0.8

0.9

1

1.1

1.2

Dimensionless strain rate p∗

Normalizedstress

Original Johnson−CookModified Johnson−Cook

Figure 5: Original vs modified Johnson-Cook model, when CJC = 0.014, CMJC = 0.013, and p0 = 5e−4.

as an infinite small value. However, it is desirable to formulate the material model as general as possible.Therefore, the first term of Eq.(40), representing the strain hardening, is replaced by another hardeningrule than that proposed by Johnson and Cook. By replacing the first term of Eq.(40) with a two termssaturation type of hardening rule proposed by Voce [15], the constitutive model could be expressed as

σy = (A+R(p)) (1 + p∗)C (1− T ∗m) = (A+

2∑k=1

Qk(1− exp (−Ckp))) (1 + p∗)C (1− T ∗m) (41)

where the constant A is the initial yield strength of the material, and Qk and Ck (k = 1, 2) define thestrain hardening. As before, the rate sensitivity is governed by the constant C, while m defines the thermalsoftening behaviour. The hardening modulus HR is found from the rate of change in R(p) given as

R =dR

dt=dR

dp

dp

dt= HR

dp

dt=⇒ HR =

dR

dp=

2∑k=1

QkCke(−Ckp) (42)

It is seen from Eq.(42) that no numerical instabilities are expected for the hardening modulus HR when theplastic strain equals zero.

Finally, evaluating yielding in the viscoplastic domain and combining Eqs.(34) and (41), the strain-ratesensitivity of the material is controlled by

p = λ =

0 for f ≤ 0

p0

[(ϕ(σ)

σy(σ0,R,T )

) 1C − 1

]for f > 0

(43)

2.3. Adiabatic thermal softening

Assuming that the plastic work is dissipated as heat implies that large plastic deformation results in anincrease in temperature. The result of this increase in temperature is a decrease in flow stress. This isalready introduced in the last term of Eq.(41), which accounts for the thermal softening of the flow stressat elevated temperatures. However, the evolution of the temperature remains to be established. Since theplastic response of the material is of a very short time duration in fast transient dynamics, the heat transferis modeled by assuming adiabatic conditions. This implies that there is no heat transfer into or out of thesystem during the plastic straining. Using the thermal energy balance per unit volume given by [10]

10

Dp = χσ : Dp = χσeqp = ρCpT (44)

where ρ is the material density, Cp is the specific heat capacity of the solid material and χ is the Taylor-Quinney coefficient. The temperature increase due to adiabatic heating can be computed as

T =Dp

ρCp(45)

From Eq.(44) it is seen that the Taylor-Quinney coefficient χ represents the fraction of the plastic power thatis converted into heat. The remaining fraction 1−χ is assumed to be stored in the material due to structuralrearrangements (e.g. elastic ”fields” around dislocations). A value of 0.9 is commonly found in the literaturefor metals, however, a conservative choice would be 1.0 which implies maximum thermal softening. In realitythe χ-value will not be constant, however, advanced experimental techniques are necessary to determine theevolution of this parameter. Moreover, one should always keep in mind that, while strain hardening (i.e.,work hardening due to creation of dislocations) implies an increase of strength locally in the material anddistributes the plasticity, softening results in localization of plasticity.

2.4. Ductile failure

A material model is not complete without some form of material degradation or failure. The degradationor failure in a material is usually given in terms of a damage parameter, and failure occurs through damageevolution. Ductile fracture arises from the nucleation, growth and coalescence of microscopic voids thatinitiate at inclusions and second phase particles. The voids around particles grow when the material issubjected to plastic straining and hydrostatic tension, and fracture occurs when the growing voids reach acritical size, relative to their spacing, resulting in a local plastic instability between the voids [16]. This workconsiders only ductile failure which is defined as the first sign of fracture, i.e., the nucleation (or initiation) ofvoids. This means that the fracture criterion is uncoupled from the constitutive equations, and is thereforedenoted a failure criterion in the following. The reader is referred to [5, 17] if damage coupling is importantin the constitutive model.

Cockcroft and Latham [9] suggested a simple failure criterion which is based on plastic work per unit volume.They reasoned that the failure criterion needed to be based on some combination of stress and strain, andthat damage accumulates during plastic straining. To account for hydrostatic tension they based the criterionon the magnitude of the major principal stress σ1, i.e.,

D =W

Wc=

1

Wc

∫ p

0

〈σ1〉 dp (46)

where Wc is the failure parameter which can be found by integrating the major principal stress in a uniaxialtension test during the entire equivalent plastic strain path until the plastic strain at failure pf . The expres-sion 〈σ1〉 is equivalent to the function max(0, σ1), which implies that only positive values of the maximumprincipal stress σ1 (i.e., tensile stress) contribute to the damage evolution. Material failure takes place whenthe damage parameter D reaches unity. This failure criterion is therefore easy to implement in a FE codeand it requires only one uniaxial tension test for calibration.

The failure criterion by Cockcroft and Latham has proven to give equally good results as more complexcriteria [18, 19, 20], and it is also shown that the criterion indirectly accounts for both deviatoric andhydrostatic stress states [21]. This is seen by using an alternative expression for the major principal stress[22], i.e.,

σ1 = σH +3 + µσ

3√

3 + µ2σ

σeq =

(σ∗ +

3 + µσ

3√

3 + µ2σ

)σeq (47)

where the stress triaxiality σ∗ and the Lode parameter µσ is defined by

11

σ∗ =σHσeq

, µσ =2σ2 − σ1 − σ3

σ3 − σ1(48)

Remember that σ1 ≥ σ2 ≥ σ3 are the ordered principal stresses. Eq.(46) can then be expressed as

D =1

Wc

∫ p

0

⟨σ∗ +

3 + µσ

3√

3 + µ2σ

⟩σeqdp (49)

Since material failure occurs when D = Dc = 1.0 and p = pf , Eq.(49) can be written as

1 =1

Wc

∫ pf

0

⟨σ∗ +

3 + µσ

3√

3 + µ2σ

⟩σeqdp (50)

Assuming that the stress triaxiality σ∗ and Lode parameter µσ remain constant throughout the entire loadinghistory (i.e., proportional loading) and using the Voce hardening rule from Eq.(41) to express the equivalentstress σeq through the yield condition in Eq.(34), Eq.(50) now reads

σ∗ =Wc

Apf +∑2k=1(Qkpf + Qk

Ck[exp (−Ckpf )− 1])

− 3 + µσ

3√

3 + µ2σ

(51)

The influence of stress triaxiality σ∗ and Lode parameter µσ on the plastic failure strain pf can then beillustrated as shown in Figure 6, where the material parameters by Holmen et al. [21] are used to generatethe respective curves. Various loading scenarios are now represented by different combinations of the stresstriaxiality and the Lode parameter. Some typical examples are (µσ, σ

∗) = (+1, 1/3) which implies uniaxialtension (σ1 ≥ σ2 = σ3), (µσ, σ

∗) = (0, 0) corresponding to pure shear (2σ2 = σ1 + σ3) and (µσ, σ∗) =

(−1,−1/3) representing uniaxial compression (σ1 = σ2 ≥ σ3).

0

1

2

3

4

5

6

7

−11/2

0

Lodepara

meter,µσ1/2

12/3Stress triaxiality, σ ∗

1/30

Plastic

failure

strain,pf

−1/3

(a) Failure surface

0

0.25

0.5

0.75

1

1.25

1.5

1.75

2

−2/3 −1/3 0 1/3 2/3Stress triaxiality, σ∗

Plastic

failure

strain,pf

µσ = −1µσ = 0µσ = +1

(b) Selected Lode parameters

Figure 6: Representation of the plastic failure strain pf in the (µσ, σ∗) space when using the Cockcroft-

Latham failure criterion and assuming proportional loading. Note that the necessary material parametersto be used in Eq. (51) are taken from Table 1.

From Eq.(46) it is seen that, to make the material model as general as possible, it is preferable to computethe principal stresses directly in the subroutine. This is done by following the procedure proposed by [12]which uses the principal deviatoric stresses σ

′

i to find the principal stresses σi. The principal deviatoric

stresses are determined from the eigenvalue problem for the deviatoric stress σ′

in Eq. (32), i.e.,

12

det(σ′I− σ′) = 0 (52)

where det(·) is the determinant of the matrix argument (·). The result of this operation is a cubic equationcalled the characteristic equation, given as

(σ′)3 − I ′1(σ

′)2 − J2σ

′ − J3 = 0 (53)

where the principal invariants are given respectively as

I′

1 = tr(σ′) = σ

′

ii = 0, J2 = −I ′2 =1

2tr(σ

′2) =1

2σ′

ijσ′

ij , J3 = I′

3 = det(σ′) (54)

and the J-symbols for the principal invariants are introduced since these are commonly used in the literaturefor metal plasticity. The general solution is then given as

σ′

1 = 2

√J2

3cos

θ

3, σ

′

2 = 2

√J2

3cos

(θ

3− 2π

3

)≤ σ′1, σ

′

3 = 2

√J2

3cos

(θ

3+

2π

3

)≤ σ′2 (55)

where the angle θ is given as

cos θ =J3

2√J3

2/27, 0 ≤ θ ≤ π (56)

and the Lode angle is recognized as θL = θ/3. Finally, the principal stresses σi of the stress tensor σ aredetermined by

σi = σ′

i + σH = σ′

i +1

3tr(σ) (57)

where the major principal stress σ1 is used in the Cockcroft-Latham (CL) failure criterion in Eq.(46).

3. Two-dimensional stress analysis

This section contains useful relationships when dealing with the particular case of plane stress. Plane stressis defined as the state of stress in which the normal stress σ33 and the shear stresses σ13,23 are assumed tobe zero [11]. From a practical point of view this assumption is typically used in the analysis of bodies inwhich one of the dimensions (i.e., the thickness) is much smaller than the others and is subjected to loadsthat result in dominant stresses perpendicular to the thickness direction [5].

3.1. Two-dimensional theory of elasticity

In the particular case of plane stress, the linear elastic stress-strain compliance relationship for an isotropicmaterial can be written as a system of equations in matrix form given as

ε =

ε11

ε22

ε33

2ε12

00

= Sσ =1

E

1 −ν −ν 0 0 0−ν 1 −ν 0 0 0−ν −ν 1 0 0 00 0 0 2(1 + ν) 0 00 0 0 0 2(1 + ν) 00 0 0 0 0 2(1 + ν)

σ11

σ22

0σ12

00

(58)

The three stress components which is assumed to be zero in the stress vector indicate that their associatedcolumns in the compliance matrix S can be neglected (i.e., column 3, 5 and 6). Furthermore, if we neglectthe rows associated with the strain components in the thickness direction εi3 (i = 1, 2, 3), the stress-straincompliance relationship reduces to a 3× 3 matrix given as

13

ε =

ε11

ε22

2ε12

= Sσ =1

E

1 −ν 0−ν 1 00 0 2(1 + ν)

σ11

σ22

σ12

(59)

Then, the elasticity matrix C for plane stress is found by inverting the plane stress compliance matrix Sand given by Hooke’s law as

σ =

σ11

σ22

σ12

= Cε =E

1− ν2

1 ν 0ν 1 00 0 1−ν

2

ε11

ε22

2ε12

(60)

It is important to note that the elasticity matrix for plane stress is not found by removing columns and rowsfrom the general isotropic elasticity matrix in Eq.(3).

Finally, in the particular case of plane stress the principal stresses (σ1, σ2) may be calculated as

σ1 =σ11 + σ22

2+

√(σ11 − σ22

2

)2

+ σ212

σ2 =σ11 + σ22

2−√(

σ11 − σ22

2

)2

+ σ212

(61)

where the major principal stress σ1 is used in the CL failure criterion in Eq.(46).

3.2. Through thickness strain increment

Although the stress in the thickness direction σ33 is assumed to be zero, this is not the case for the corre-sponding strain ε33. The expression for the change in thickness during plane stress may be found by usingthe relationship between the volumetric strain εv(= εkk) and the hydrostatic pressure P from the theory ofthree-dimensional elasticity, i.e.,

εv = εkk = − 1

KP, K = λel +

2

3µel =

E

3(1− 2ν)(62)

where P = − 13σkk is the mean stress and K is the bulk modulus. Note that the hydrostatic pressure P is

equal to the hydrostatic stress σH but has the opposite sign.

The physical meaning of εkk is the relative change of volume

εkk = ε11 + ε22 + ε33 =dV

V(63)

where experimental evidence has shown that plastic deformation is volume preserving (dV = 0) even at highhydrostatic pressures [10]. This means that εkk = 0 and that the material is incompressible during plasticdeformation. Thus, the strain through the thickness for plates and shells is given as

ε33 = − (ε11 + ε22) (64)

where it is seen that the strain in the thickness direction ε33 is a result of the in-plane strains ε11,22. This istypical for thin-walled structures where there are no constraints for the strain in the thickness direction. Theonly inconsistency is that in the constitutive equations for plates and shells the thickness is considered tobe constant while Eq.(64) shows that in reality there will be a small change. This illustrates the importanceof defining the thickness strain increment ∆ε33 in the material routine to predict the actual change in thethickness of the element.

14

Using the assumption of a plastic incompressible material (i.e., εpkk = 0) the plastic strain increment throughthe thickness (due to plastic thinning) is given as

∆εp33 = −(∆εp11 + ∆εp22) (65)

where ∆εp11 and ∆εp22 could be expressed using strain decomposition

∆εp11 = ∆ε11 −∆εe11

∆εp22 = ∆ε22 −∆εe22

(66)

and the elastic strain components are given by Hooke’s law in Eq.(59)

∆εe11 =1

E(∆σ11 − ν∆σ22)

∆εe22 =1

E(∆σ22 − ν∆σ11)

(67)

By combining Eqs.(65) to (67), the plastic strain increment through the thickness is given by

∆εp33 = −(∆ε11 −∆εe11)− (∆ε22 −∆εe22)

= ∆εe11 + ∆εe22 −∆ε11 −∆ε22

=1

E(∆σ11 − ν∆σ22 + ∆σ22 − ν∆σ11)− (∆ε11 + ∆ε22)

=1− νE

(∆σ11 + ∆σ22)− (∆ε11 + ∆ε22)

(68)

Then, when the plastic strain increment ∆εp33 is known the total through thickness strain increment ∆ε33

is finally found by strain decomposition

∆ε33 = ∆εe33 + ∆εp33

= − νE

(∆σ11 + ∆σ22) +1− νE

(∆σ11 + ∆σ22)− (∆ε11 + ∆ε22)

=1− 2ν

E(∆σ11 + ∆σ22)− (∆ε11 + ∆ε22)

(69)

4. Uniaxial stress case

The case of uniaxial stress, which occurs in bars and in some types of beams, can be treated similarly to thetwo-dimensional stress analysis case presented in Section 3.1.

In this case one assumes that the two lateral normal stresses are zero (σ22 = σ33 = 0), so that the corre-sponding strain increments ∆ε22 and ∆ε33 must be computed and returned by the consitutive routine.

We assume that the strain increments can be decomposed into elastic and plastic parts, so that

∆εp11 = ∆ε11 −∆εe11

∆εp22 = ∆ε22 −∆εe22 (70)

∆εp33 = ∆ε33 −∆εe33

and that the plastic part of the deformation occurs at constant volume (dV = 0), due to plastic incompress-ibility

∆εp11 + ∆εp22 + ∆εp33 = 0 (71)

15

In the general 3D case, the elastic strain increments are related to the total stress increments by Hooke’slaw

∆εe11 =1

E[∆σ11 − ν (∆σ22 + ∆σ33)]

∆εe22 =1

E[∆σ22 − ν (∆σ11 + ∆σ33)] (72)

∆εe33 =1

E[∆σ33 − ν (∆σ11 + ∆σ22)]

However, in the uniaxial stress case the lateral normal stresses are assumed to vanish (σ22 = σ33 = 0) andtherefore also the stress increments vanish (∆σ22 = ∆σ33 = 0). Eq. (72) then reads

∆εe11 =1

E∆σ11

∆εe22 = − νE

∆σ11 = −ν∆εe11 (73)

∆εe33 = − νE

∆σ11 = −ν∆εe11

which implies that

∆εe22 = ∆εe33 = −ν∆εe11 (74)

and by utilizing the symmetry it is also assume that

∆εp22 = ∆εp33 (75)

and therefore

∆ε22 = ∆ε33 (76)

ε22 = ε33 (77)

Then, by replacing Eq. (75) into Eq. (71) one obtains

2∆εp22 = 2∆εp33 = −∆εp11 (78)

∆εp22 = ∆εp33 = −1

2∆εp11 = −1

2(∆ε11 −∆εe11) (79)

Using Eq. (73) this now becomes

∆εp22 = ∆εp33 = −1

2

(∆ε11 −

1

E∆σ11

)= −1

2∆ε11 +

1

2E∆σ11 (80)

Finally, the total lateral strain increment ∆ε22 (and ∆ε33) is obtained by strain decomposition

∆ε22 = ∆ε33 = ∆εe22 + ∆εp22 = ∆εe33 + ∆εp33

= −1

2∆ε11 +

1

2E∆σ11 − ν∆εe11

=1

2E∆σ11 −

ν

E∆ε11 −

1

2∆ε11

=1− 2ν

2E∆σ11 −

1

2∆ε11

(81)

16

5. Numerical Return Mapping

Wilkins [23] was one of the first to introduce a return mapping procedure, and a variety of schemes havebeen proposed following this work (see e.g. [2, 5, 4, 3, 24]). In general, it is possible to categorize thestress integration methods in two different approaches, i.e., the forward Euler method and the backwardEuler method. The choice of method is illustrated in Figure 7 and is mainly based on the assumption ofthe direction of the plastic flow n = ∂f/∂σ. The forward Euler method uses the known stress state at theprevious or trial configuration to determine the plastic flow direction n [2] (see Figure 7a). This assumptionis therefore only valid for very small time steps, and is therefore suitable for the explicit time integrationFE method. Since the direction is known there is only one unknown during the return mapping procedure,i.e., the plastic multiplier λ. The backward Euler method evaluates the normal n to the yield surface at thecurrent unknown stress state [3] (see Figure 7b), which gives good accuracy even for larger time steps makingit appropriate for both the explicit and implicit time integration FE methods. However, the variation of thenormal to the yield surface must be taken into consideration during the return mapping procedure resultingin a more complex integration, due to the involvement of possible 2nd derivatives of the yield function. Inthe particular case of isotropic von Mises plasticity, the accuracy of the forward and backward Euler methodis found to be equal [25]. It should also be emphasized that when considering viscoplasticity the plasticstrain rate will change during the local iterations in the forward Euler method. That is, the plastic strainrate will increase during the iterative return mapping (due to an increase in plastic strain). This impliesthat the return to the yield surface (f = 0) occurs at strain rates that are too low. However, the plasticstrain rate at the converge increment ∆tn+1 should be the same as in the backward Euler method.

σn

σtrial

n+1

σ2n+1

σ1n+1

σn+1

fn+1 = 0

(a) Explicit stress update using forward Euler(cutting plane method)

σn

σtrial

n+1

σ2n+1

σ1n+1

σn+1

fn+1 = 0

(b) Fully implicit stress update using backward Euler(closest point projection method)

Figure 7: The elastic predictor viscoplastic corrector return map algorithms for the von Mises yield criterionand associated plasticity. Starting from an elastic stress state at the last converged increment tn beforearriving at the new converged increment tn+1.

During both approaches an elastic predictor uses the total strain increment and the linear elastic relationto obtain a trial stress state. Then, if the elastic trial stress is outside the plastic domain, the elasticallypredicted stresses are corrected to fit a suitably updated yield surface by iterations of the plastic corrector.Thus, the proposed integration algorithms are formulated as an elastic predictor viscoplastic corrector stressupdate using a return mapping scheme. The solution is strain controlled in the sense that the total (co-rotational) strain increment ∆εn+1 = Dn+1∆tn+1, within the time interval ∆tn+1 = tn+1 − tn, is knownfrom the global equilibrium and used to update the stresses and internal variables at step n+ 1. Then, as in

17

a rate-independent case, it makes sense to first compute an elastic trial state by assuming that the materialbehaviour is purely elastic within the interval. If the trial state is within the elastic domain (f(σ) ≤ 0),no viscoplastic flow takes place within the considered time step and the trial state is the actual state atthe end of the step. Otherwise, the evolution of stresses and internal variables is computed by means of asuitable return mapping method. It should be emphasized that the consistency condition (f(σ) = 0) nolonger holds in the viscoplastic case, since the updated stress state at tn+1 generally lies outside the yieldsurface (f(σ) > 0). However, the terminology viscoplastic return mapping is justified in the present casesince the updated stresses are obtained by returning (or moving) the trial stress towards the yield surface.Hence, the application of the procedure is similar to that in the rate-independent case, and Eq. (23) reads

∆σ = C : (∆ε−∆εp) = ∆σe + ∆σp (82)

where the stress update at time tn+1 is expressed as

σn+1 = σn + ∆σn+1 = σn + C : ∆εn+1 −∆λn+1C : nn+1 = σtrialn+1 −∆σp (83)

in which the viscoplastic corrector ∆σp returns the stress to the yield surface.

All internal variables and functions in Eqs.(20), (23), (34), (36), (42), (43), (45) and (46) are thereforeadopted to the return mapping scheme for integration, and replaced with the corresponding incrementalvalues within and at the end of the considered interval ∆tn+1. The incremental form of the system of firstorder differential equations then reads

εn+1 = εn + ∆εn+1

εpn+1 = εpn + ∆εpn+1 = εpn + ∆λn+1nn+1 = εpn +3

2∆λn+1

σ′

n+1

σeq,n+1

λn+1 = λn + ∆λn+1 = λn + p0∆tn+1

[(ϕn+1

σy,n+1

) 1C

− 1

]σn+1 = σn + ∆σn+1 = σn + C : (∆εn+1 −∆εpn+1) = σtrialn+1 −∆λn+1C : nn+1

Rn+1 = Rn + ∆Rn+1 = Rn + ∆λn+1HR,n+1 = Rn + ∆λn+1

2∑k=1

QkCkexp(−Ckλn+1)

σy,n+1 = σ0 +Rn+1

Tn+1 = Tn + ∆Tn+1 = Tn +χσeq,n+1∆λn+1

ρCp

Dn+1 = Dn + ∆Dn+1 = Dn +〈σ1,n+1〉∆λn+1

fv,n+1 = ϕn+1 − σy,n+1

(1 +

∆λn+1

p0∆tn+1

)C (1−

[Tn+1 − TrTm − Tr

]m)

(84)

where ∆λn+1 is the incremental plastic multiplier at time tn+1.

The set of nonlinear algebraic equations in Eq. (84) are solved by updating the normal to the yield surfacegradient nn+1 iteratively, and the viscoplastic corrector is applied from the elastic predictor σtrialn+1 throughoutthe iteration until fv,n+1 = 0.

5.1. Cutting plane method

The present study focus mainly on the cutting plane method originally proposed by Ortiz and Simo [2]. Thebasis for this return mapping is an explicit elastic predictor plastic corrector stress update using a forwardEuler scheme. That is, the objective of the numerical integration is to use the known stress state at the

18

previous or trial configuration n to determine the new converged stress state at step n+ 1 (Figure 7a).

By using the normal nin+1 at the previous stress state (see Figure 7a) and assuming weak coupling of thetemperature, i.e., using Tn instead of Tn+1, there is only one unknown during the return mapping procedureand that is the plastic multiplier ∆λi+1

n+1. This reduces the fully set of equations in Eq. (84) to a singleequation which can be solved by a linear Taylor-expansion around fn+1(σ,X) = fn+1(σ, σy, υp(p),Γp(Tn)),i.e.,

f i+1n+1 = f in+1 +

∂f

∂σ

∣∣∣∣in+1

∆σn+1 +∂f

∂X

∣∣∣∣in+1

∆Xn+1 (85)

which may be written in terms of the internal variables X as

f i+1n+1 = f in+1 +

∂f

∂σn+1δσn+1 +

∂f

∂σy,n+1δσy,n+1 +

∂f

∂υp,n+1δυp,n+1 = 0 (86)

where the flow stress σy,n+1 and the following expressions are chosen as the internal variables X in theTaylor-expansion

υp(∆λn+1) =

(1 +

∆λn+1

p0∆tn+1

)CΓp(Tn) =

(1−

[Tn − TrTm − Tr

]m)(87)

Introducing the partial derivatives of the internal variables in Eq. (86) this now reads

f i+1n+1 = f in+1 −

∂f

∂σn+1C

∂f

∂σn+1δλi+1

n+1 − υp,n+1Γp,nHR,n+1δλi+1n+1 − σy,n+1Γp,n

∂υp,n+1

∂λn+1

δλi+1n+1 = 0 (88)

Moreover, using p = λ = ∆λ∆t the internal variable controlling the viscoplasticity υp(p) may be expressed as

υp(∆λ) and Eq. (88) may be written as

f i+1n+1 = f in+1 −

∂f

∂σn+1C

∂f

∂σn+1δλi+1

n+1 − υp,n+1Γp,nHR,n+1δλi+1n+1 − σy,n+1Γp,n

∂υp,n+1

∂∆λn+1δλi+1

n+1 = 0 (89)

Finally, the incremental change δλ in the incremental plastic multiplier ∆λ is then expressed as

δλi+1n+1 =

f in+1

∂f∂σn+1

C ∂f∂σn+1

+ υp,n+1Γp,nHR,n+1 + σy,n+1Γp,n∂υp,n+1

∂∆λn+1

(90)

where

∂υp(∆λn+1)

∂∆λ=

C

p0∆tn+1

(1 +

∆λn+1

p0∆tn+1

)C−1

(91)

HR,n+1 =dRn+1

dpn+1=

2∑k=1

QkCkexp(−Ckλn+1) (92)

∂f

∂σ

T[C]

∂f

∂σ

=

∂f

∂σijCijkl

∂f

∂σkl= C1111

[(∂f

∂σ11

)2

+

(∂f

∂σ22

)2

+

(∂f

∂σ33

)2]

+ 2C1122

[∂f

∂σ11

∂f

∂σ22+

∂f

∂σ22

∂f

∂σ33+

∂f

∂σ33

∂f

∂σ11

]+ 4C1212

[(∂f

∂σ12

)2

+

(∂f

∂σ23

)2

+

(∂f

∂σ31

)2] (93)

19

For more details regarding the linear Taylor-expansion, it is referred to Appendix A. It is also emphasizedthat since the time increment ∆tn+1 is often very small during explicit time integration, the change in tem-perature ∆Tn+1 is assumed to have a negligible effect on the update of the state variables. Therefore, thetemperature is assumed to have no evolution during the time increment ∆tn+1, and a low coupling approachfor the temperature is applied. This implies that Tn is used in the return mapping to update the stressand internal variables at time tn+1, and then the resulting state variables at tn+1 are used to calculate andupdate Tn+1. Thus, this assumption simplifies the calculations leading to Eq. (90) without any significantloss of accuracy in the model. This is also the case for the normal nn+1 = ∂f/∂σn+1 to the yield surfacewhere the use of the previous normal nin+1 instead of ni+1

n+1 violates the associated flow rule in Eqs. (36) and(84) since a unique normal for the given strain increment ∆εn+1 must be found at each stress state. Thus,the deformation does not follow the minimum plastic work path when the cutting plane method is used (seeFigure 7). However, when the time increment ∆tn+1 is very small this is assumed to have negligible effecton the stress state.

The numerical scheme within the time increment ∆tn+1 may then be summarized as follows (see alsoAppendix E):

1. Set the initial values of internal variables to the converged values of the previous step at tn, check ifthe integration point under evaluation has already failed, and calculate the speed of sound.

2. Compute the elasticity tensor.That is, IF a 3D stress state is used THEN use Eqs. (3) and (4)

C = λelI⊗ I + 2µelI =

C1 C4 C4 0 0 0C4 C1 C4 0 0 0C4 C4 C1 0 0 00 0 0 C7 0 00 0 0 0 C7 00 0 0 0 0 C7

C1 =

(1− ν)E

(1 + ν)(1− 2ν), C4 =

νC1

1− ν , C7 =1

2

E

1 + ν

ELSEIF a 2D stress state is used THEN use Eq. (60)

C =

C1 C4 0C4 C1 00 0 C7

, C1 =E

1− ν2, C4 = νC1, C7 =

1

2

E

1 + ν

ELSE a 1D stress state is used THEN use

C =

C1 0 00 0 00 0 0

, C1 = E

3. Use the total stain increment ∆εn+1 from the global equilibrium to compute the elastic predictor

σtrialn+1 = σn + C : ∆εn+1

4. Compute the von Mises equivalent stress in terms of the elastic trial stress and Eq. (35)

σtrialeq,n+1 = ϕ(σtrialn+1 ) =

√3

2σ′trialn+1 : σ

′trialn+1

20

5. Check if temperature softening is included in the material input (i.e., m 6= 0).

6. Set the incremental plastic multiplier ∆λn+1 equal to zero.

7. Check for plastic admissibility, i.e., IF f(σtrialn+1 ) > 0 THEN apply return mapping by using the cuttingplane algorithm (SOLU =1) to find the incremental change in the plastic multiplier ∆λn+1. Note thatsuperscript i denotes the local iteration counter in the following.

(i) Compute the normal nn+1 to the yield surface based on the initial (i.e., trial) or previous stressconfiguration.

nn+1 =3

2

σ′

n+1

σeq

(ii) Compute hardening modulus HR,n+1 according to Eq. (42)

HiR,n+1 = Q1C1 exp(−C1p

in+1) +Q2C2 exp(−C2p

in+1)

(iii) Compute denominator in Eq. (90) using Eqs. (91), (92), (93) and (98)

DENOM =∂f

∂σin+1

C∂f

∂σin+1

+ υip,n+1Γp,nHiR,n+1 + σiy,n+1Γp,n

∂υip,n+1

∂∆λin+1

(iv) Compute incremental change in the plastic multiplier δλi+1n+1 in Eq. (90) and update variable

∆λi+1n+1

δλi+1n+1 =

f in+1

∂f∂σn+1

C ∂f∂σn+1

+ υp,n+1Γp,nHR,n+1 + σy,n+1Γp,n∂υp,n+1

∂∆λn+1

∆λi+1n+1 = ∆λin+1 + δλi+1

n+1

(v) Update internal variables dependent on ∆λi+1n+1

pi+1n+1 = pin+1 + δλi+1

n+1

Ri+1n+1 = Rn +Hi

R,n+1∆λi+1n+1

σi+1y,n+1 = A+Hi+1

R,n+1∆λi+1n+1

(vi) Update stress components using the elastic Hooke’s law according to Eq. (83)

σi+1n+1 = σtrialn+1 −∆λi+1

n+1C : ni+1n+1

Note that the material routine uses a a fully vectorized representation of the stress and strainvectors, i.e.,

σ = [σ11, σ22, σ33, σ12, σ23, σ31]T

ε = [ε11, ε22, ε33, γ12, γ23, γ31]T = [ε11, ε22, ε33, 2ε12, 2ε23, 2ε31]T

Moreover, the application of objective stress rates (i.e., the Jaumann rate in EPX) to update thestress tensor σ in plane stress states allow for the possibility that the normal stress σ33 will notbe zero (outside the material routine). The material subroutine should therefore initialize thenormal stress σ33 to zero (σ33 = 0).

(vii) Compute the updated von Mises equivalent stress using Eq. (35)

σi+1eq,n+1 = ϕ(σi+1

n+1) =

√3

2σ′,i+1n+1 : σ

′,i+1n+1

21

(viii) Update the yield function in Eq. (84) according to the new stress state computed in (vi) and (vii)

f i+1v,n+1 = ϕi+1

n+1 − σi+1y,n+1

(1 +

∆λi+1n+1

p0∆tn+1

)C (1−

[Tn − TrTm − Tr

]m)(viiii) Check for convergence, i.e., IF ∣∣∣∣∣ fv(∆λi+1

n+1)

σy,n+1υpΓp

∣∣∣∣∣ < TOL

THEN continue to 7.ELSE i = i+ 1 and GO TO (i)

8. Update the temperature according to the low coupling approach using Eq. (45) and check IF T ≥ Tm

Tn+1 = Tn +χσeq,n+1∆λn+1

ρCT

9. Compute the hydrostatic and deviatoric stress respectively according to Eq. (32)

σH,n+1 =1

3tr(σn+1)

σ′

n+1 = σn+1 − σH,n+1I

10. Computing the principal stresses σi,n+1 based on the principal deviatoric stresses σ′

i,n+1 as shown inEqs. (55) and (57)

σ′

1,n+1 = 2

√J2,n+1

3cos

θ

3

σ′

2,n+1 = 2

√J2,n+1

3cos

(θ

3− 2π

3

)≤ σ′1,n+1

σ′

3,n+1 = 2

√J2,n+1

3cos

(θ

3+

2π

3

)≤ σ′2,n+1

where

cos θ =J3,n+1

2√J3

2,n+1/27, 0 ≤ θ ≤ π

σi,n+1 = σ′

i,n+1 + σH,n+1 = σ′

i,n+1 +1

3tr(σn+1)

11. Compute the damage parameter Dn+1 by using Eq. (46), i.e., integrating the major principal stressduring the entire equivalent plastic strain path

Wn+1 = Wn + max(σ1,n+1, 0)∆λn+1

Dn+1 =Wn+1

Wc

12. Update state variables to be returned to the FE code and check for element erosion, i.e., IF Dn+1 > DC

THEN the material point stiffness is deleted.

22

13. Update the (actual) total strain ε33,n+1 and incremental strain ∆ε33,n+1 through the thickness accord-ing to Eq. (69) and return this to the FE code

∆ε33,n+1 =1− 2ν

E(∆σ11,n+1 + ∆σ22,n+1)− (∆ε11,n+1 + ∆ε22,n+1)

ε33,n+1 = ε33,n + ∆ε33,n+1

In a FE code, this is typically done for each material integration point in the FE assembly. The stresses areupdated in the material interface using a fully vectorized version of the forward Euler integration algorithmand a two-state architecture where the initial values at tn are stored in the old arrays and the new valuesat tn+1 must be updated and stored in the new arrays returned to the global FE analysis. The VPJCmaterial routine is valid for 1D, 2D and 3D stress states, i.e., for bar, shell, solid, axisymmetric, plane strainand plane stress elements. The stresses for 3D elements are stored similar to that of symmetric tensorsas σ = [σ11, σ22, σ33, σ12, σ23, σ31]T , and plane stress, axisymmetric and plane strain elements are stored asσ = [σ11, σ22, σ33, σ12]T . The deformation gradient and strains are stored similarly to the stresses, however,one should be aware of that the shear strain is stored as engineering shear strains, e.g. γ12 = 2ε12.

5.2. Radial return method

The objective of the implicit backward Euler integration algorithm is to find the minimum distance fromthe trial state to the final state at step n+ 1 (see Figure 7b). In the particular case of a 3D stress state, vonMises yield criterion with associated flow rule and isotropic hardening, the fully implicit set of equations inEq. (84) is reduced to a single equation and the radial return algorithm. This is due to the circular cylindricalshape of the yield surface in Figure 4a, which implies that the gradient of the yield surface at the trial statentrialn+1 is co-linear with the corresponding gradient at the final stress state nn+1 (Figure 8). The return tothe yield surface from the elastic trial state is radial in the deviatoric plane, which simplifies the algorithmand makes the return mapping exceptionally stable and accurate. This makes the radial return mapping oneof the most widely used integration algorithms for plain strain and 3D J2 (or von Mises) elastoplasticity [26].

By considering the deviatoric stress space, the equivalent stress could then be expressed as (see also AppendixB)

σeq = ϕ(σ′

n+1) = ϕ(σ′trialn+1 )− 3G∆λn+1, G =

E

2(1 + ν)(94)

and the system of equations in Eq. (84) is only dependent on the incremental plastic multiplier ∆λn+1,where the viscoplastic constraint fv,n+1(σ) = 0 reads

fv,n+1(σ) = ϕtrialn+1 (σ)− 3G∆λn+1 − σy,n+1

(1 +

∆λn+1

p0∆t

)C (1−

[Tn+1 − TrTm − Tr

]m)= 0 (95)

Then, the incremental plastic multiplier ∆λn+1 can be solved by means of a local Newton-Raphson iteration,i.e.,

∆λi+1n+1 = ∆λin+1 −

fv(∆λin+1)

∂fv(∆λin+1)

∂∆λ

(96)

where (again) i is the local iteration counter.

Since the time increment ∆tn+1 is often very small during explicit time integration, the change in temperature∆Tn+1 is assumed to have a negligible effect on the update of the state variables. Hence, the temperatureis assumed to have no evolution during the time increment ∆tn+1, and a low coupling approach for thetemperature is applied. This implies that Tn is used in the return mapping to update the stress and internal

23

σ3

σ2σ1

π-plane ∂f∂σ

∣∣∣n+1

σ′

n

σ′trialn+1

σ′

n+1

3G∆λ n

+1

ϕ(σ′ trial

n+

1)σ′ trial

n+

1

surface at tn

surface at tn+1

Figure 8: The implicit elastic predictor viscoplastic corrector return map algorithm for the von Mises yieldcriterion and associated plasticity in the deviatoric plane.

variables at time tn+1, and then the resulting state variables at tn+1 are used to calculate Tn+1. Thisassumption reduces the number of equations in Eq. (84), and the residual derivative in Eq. (96) is expressedas

∂fv(∆λn+1)

∂∆λ=

∂

∂∆λ

[ϕ(σ′trialn+1

)− 3G∆λn+1 − σy,n+1υp(∆λn+1)Γp(Tn)

]= −3G− ∂σy,n+1

∂∆λυp(∆λn+1)Γp(Tn)− σy,n+1

∂υp(∆λn+1)

∂∆λΓp(Tn)

(97)

where

υp(∆λn+1) =

(1 +

∆λn+1

p0∆t

)C∂υp(∆λn+1)

∂∆λ=

C

p0∆t

(1 +

∆λn+1

p0∆t

)C−1

Γp(Tn) =

(1−

[Tn − TrTm − Tr

]m) (98)

The numerical scheme within the time increment ∆tn+1 may then be summarized as follows:

1. Set the initial values of internal variables to the converged values of the previous step at tn, and setthe incremental plastic multiplier ∆λn+1 equal to zero. Use the total stain increment ∆εn+1 from theglobal equilibrium to compute the elastic predictor

σtrialn+1 = σn + C : ∆εn+1

2. Compute the equivalent stress in terms of the elastic trial stress

σtrialeq = ϕ(σ′trialn+1 ) =

√3

2σ′trialn+1 : σ

′trialn+1

and the radial normal vector for the yield surface

nn+1 =3

2

σ′trialn+1

σtrialeq

24

3. Check for plastic admissibility, i.e., IF f(σ′trialn+1 ) > 0 THEN perform local Newton-Raphson iteration

to find ∆λn+1

(i) ∆λi+1n+1 = ∆λin+1 −

fv(∆λin+1)

∂fv(∆λin+1

)

∂∆λ

(ii) IF

∣∣∣∣ fv(∆λi+1n+1)

σy,n+1υpΓp

∣∣∣∣ > TOL THEN i = i+ 1 and GO TO (i)

4. Update internal variables dependent on ∆λn+1. Note that only the deviatoric stress will change duringthe radial return, and that the hydrostatic stress could be calculated based on the elastic trial stress.

5. Update stress components using the elastic Hooke’s law, i.e.,

σn+1 = σtrialn+1 −∆λn+1C : nn+1

6. Check for element erosion, i.e., IF D > DC THEN the material point stiffness is deleted.

7. Update state variables to be returned to the global analysis in EPX.

As for the cutting plane method in Section 5.1, this is done for each material integration point in the FEassembly and the stresses are updated using a two-state architecture and returned to the global analysis inEPX. It must be emphasized that the radial return method is only valid in the case of a 3D stress state(Figure 4), and therefore limited to solid, axisymmetric and plane strain elements. The stresses for three-dimensional elements are stored similar to that of symmetric tensors as σ = [σ11, σ22, σ33, σ12, σ23, σ31]T , andplane strain and axisymmetric elements are stored as σ = [σ11, σ22, σ33, σ12]T . The deformation gradientand strains are stored similarly to the stresses, however, one should be aware of that the shear strain isstored as engineering shear strains, e.g. γ12 = 2ε12.

6. Verification of Material Subroutine

The performance of the material subroutine is verified and validated in EPX. First, a verification is per-formed using single element simulations in uniaxial tension and simple shear. Then, the model is validatedagainst experimental data found in the literature for quasi-static tension tests [21] and airblast experiments[27].

The material model requires the user to specify 20 parameters, where 16 are mandatory while the room (orambient) temperature Tr, convergence criterion TOL, maximum number of local iterations MXITER andpreferred solution algorithm SOLU are taken as default if not specified in the input file [7]. The materialconstants used for the verification and validation in the following are taken from the literature for a ductilestructural steel [21] and summarized in Table 1. Previous studies by De Borst and Feenstra [25] have shownthat the choice of return mapping algorithm is more important for orthotropic yield criteria (e.g. the Hillcriterion) than for the isotropic von Mises plasticity theory used in this study. The cutting-plane algorithm(SOLU = 1) is the default choice and will therefore be used for the return mapping because this is robustand readily implemented for all stress states (1D, 2D and 3D).

6.1. Singel element verification

An extensive single element verification of the VPJC material model is performed in Reference [28]. Thefollowing single element verification is therefore mainly included for completeness of this study, and it isreferred to [28] for more detailed single element verifications. Since the material model is applicable for one-(1D), two- (2D) and three-dimensional (3D) stress states, the scope of this work is limited to one represen-tative bar element in 1D (FUN3 ), one continuum element in 2D (Q42L), one shell element in 3D (Q4GS )and one continuum element in 3D (CUBE ). This is based on the findings in [28], which concluded that the

25

Table 1: Material parameters used in numerical simulations [7, 21].

Elastic constants Yield stress Strain-rate Damageand density and strain hardening hardening

E ν ρ A Q1 C1 Q2 C2 p0 C WC DC

[MPa] [ - ] kg/m3 [MPa] [MPa] [ - ] [MPa] [ - ] [s−1] [ - ] [MPa] [ - ]210000 0.33 7850 370 236.4 39.3 408.1 4.5 5e-4 0.01 473 1.0

Adiabatic heating and temperature softening Convergence Solution algorithmCp χ Tm Tr m TOL MXITER SOLU

[J/kgK] [ - ] [K] [K] [ - ] [ - ] [ - ] [1: cutting plane or 2: radial return]452 0.9 1800 293 1.0 10−5 100 1

results are more or less independent of element type. Note that the JAUM option must be activated whenusing CEA continuum elements (e.g. Q42L and CUBE ) to treat large strains, and that the LMST optionmust be included to update the thickness for the Q4GS elements during plastic straining [7].

The numerical tests of each element are performed in uniaxial tension and simple shear for the 2D and 3Delements, while the 1D element is limited to uniaxial tests only. Note that, in addition to tests in uniaxialtension and simple shear, it is common to validated a material subroutine in terms of uniaxial compression,uniaxial tension/compression in oblique directions, and uniaxial tension/compression with a finite rotation.However, since the formulation of the material model is isotropic the response is the same in tension andcompression and independent of loading direction. It is therefore considered sufficient to only test one ofthese loading conditions. Moreover, since the co-rotational formulation in the material routine is taken careof by EPX outside the material subroutine environment [7], there is no need to test the return map algorithmin terms of finite rotation. The verification of the material routine is therefore limited to the uniaxial tensionand simple shear tests in Figure 9.

∆L

L0

v3 = L

(a) Uniaxial tension

θ

v2 = u2u2

L0

(b) Simple shear

Figure 9: Illustration of the loading cases for the single element tests of the brick element (CUBE ). Notethat γ = tan(θ) = u2/L0.

26

The numerical models consist of a simple input file containing one element with element lengths of L0 = 1mm. The tests are performed with a prescribed displacement resulting in an equivalent plastic strain ofapproximately 2.0. This is obtained by applying a constant velocity throughout the analysis (Figure 9), andthe results are compared with the analytical solution for the equivalent stress in Eq. (41). In the particularcase of including viscoplasticity (C 6= 0), it is convenient to test the elements at constant strain rate sincethe analytical solution of the equivalent stress is readily obtained from Eq. (41). Due to the logarithmicnature of the longitudinal strain in uniaxial tension, i.e.,

ε33 = εe33 + εp33 = ln

(L

L0

)= ln

(1 +

∆L

L0

)(99)

It is necessary with an exponential displacement-driven loading to obtain a constant strain rate. By consid-ering large strains and neglecting the elastic part εe33 of the longitudinal strain, Eq. (99) may be simplifiedto

ε33 ≈ εp33 = p = pt = ln

(L(t)

L0

)(100)

which may be expressed as

L(t) = L0 exp (pt) (101)

where the displacement is given by

d33(t) = L(t)− L0 = L0 (exp (pt)− 1) (102)

with corresponding velocity

v3(t) = L0p exp (pt) , v3(t = 0) = v3,0 = L0p (103)

Eq. (102) was therefore used to ensure constant strain rate in uniaxial tension, by choosing typical valuesof the plastic strain rate p and controlling the respective termination times such that the final displacementwas the same in all tests. The corresponding viscoplastic response in the simple shear tests was obtained byvarying the constant velocity (v2 in Figure 9) and termination time to obtain an equivalent plastic strainof approximately 2.0. To compare the tests in simple shear with the analytical solution in Eq.(41), it wasnecessary to scale the loading velocity in simple shear (v2 = constant) with respect to the initial velocity inuniaxial tension (v3(t = 0) = v3,0). The scaling factor was obtained by considering the plastic strain rate pdefined in Eq. (37), which for the uniaxial test reads

p =

√2

3Dp : Dp =

√2

3

[(εp11)

2+ (εp22)

2+ (εp33)

2]

=

√2

3

[6

4(εp33)

2

]= εp33 (104)

when assuming that the material is incompressible during plastic deformation (see Eq. (64)). The corre-sponding expression in simple shear is

p =

√2

3Dp : Dp =

√√√√2

3

[(1

2˙γ12

)2

+

(1

2˙γ21

)2]

=˙γ12√3

=γ√3

(105)

This implies that the loading velocity in simple shear was v2 =√

3v3. In addition, the termination timein the case of simple shear was scaled such that tests in uniaxial tension and simple shear had the sameequivalent plastic strain at the end of the simulation (see Table 2).

To illustrate the capabilities of the model, the model is first tested with respect to rate-independent plas-ticity. Then, the model is tested for its precision in predicting both viscoplastic and thermoviscoplastic

27

Table 2: Loading in numerical test programme.

Loading Uniaxial tension Simple shearVelocity [m/s] v3,0 = 1.0 0.1 0.01 0.001 v2 = 1.732 0.1732 0.01732 0.001732Termination [s] 0.002 0.02 0.2 2 0.0021 0.021 0.21 2.1

p [s−1] 1000 100 10 1 1000 100 10 1

behaviour, before finally tested with respect to ductile failure. The numerical results were compared to ana-lytical solutions for the constitutive equations, where this was available, to ensure that the overall behaviourof the material model is correct.

6.1.1. Rate-independent plasticity

The rate-independent behaviour of the material model is verified by excluding the strain-rate and temper-ature sensitivity. This is readily done in the material input by choosing C = 0 and m = 0. The damagethreshold was also set to an exaggerated value to avoid element erosion during plastic straining, i.e., DC 1.The analytical solution then reduces to the first term in Eq. (41), which will be used for comparison to thenumerical solution.

1D bar element

Figure 10 shows the performance of the model in describing the rate-independent behaviour in uniaxialtension for a bar element (FUN3 ).

0 0.5 1 1.5 20

200

400

600

800

1000

1200

Plastic strain p [mm/mm]

Str

ess [

MP

a]

σeq

(EPX)

R(p) (EPX)

σeq

(analytical)

R(p) (analytical)

Figure 10: Results from strain-rate and temperature independent 1D single element (FUN3 ) test in uniaxialtension (C = 0, m = 0 and DC 1). Numerical results in EPX compared to analytical solution in Eq. (41).

2D continuum element

Figure 11 shows the capability of the model in describing the rate-independent behaviour in uniaxial tensionand simple shear for a 2D continuum and plane stress condition by using the fully integrated quadrilateralQ42L element. Note that the JAUM option must be activated when using CEA continuum elements (e.g.Q42L) to treat large strains.

28

0 0.5 1 1.5 20

200

400

600

800

1000

1200

Plastic strain p [mm/mm]

Str

ess [

MP

a]

σeq

(EPX)

R(p) (EPX)

σeq

(analytical)

R(p) (analytical)

(a) Uniaxial tension (Q42L)

0 0.5 1 1.5 20

200

400

600

800

1000

1200

Plastic strain p [mm/mm]

Str

ess [

MP

a]

σeq

(EPX)

R(p) (EPX)

σeq

(analytical)

R(p) (analytical)

(b) Simple shear (Q42L)

Figure 11: Results from strain-rate and temperature independent 2D single element (Q42L) tests (C = 0,m = 0 and DC 1). Numerical results in EPX compared to analytical solution in Eq. (41).

3D shell element

The same tests in uniaxial tension and simple shear are performed also for a 3D (quadrilateral) shell element(Q4GS ), where the numerical results are compared to the analytical solution from Eq. (41) in Figure 12.Note that the LMST option must be included to update the thickness for the Q4GS elements during plasticstraining.

0 0.5 1 1.5 20

200

400

600

800

1000

1200

Plastic strain p [mm/mm]

Str

ess [

MP

a]

σeq

(EPX)

R(p) (EPX)

σeq

(analytical)

R(p) (analytical)

(a) Uniaxial tension (Q4GS)

0 0.5 1 1.5 20

200

400

600

800

1000

1200

Plastic strain p [mm/mm]

Str

ess [

MP

a]

σeq

(EPX)

R(p) (EPX)

σeq

(analytical)

R(p) (analytical)

(b) Simple shear (Q4GS)

Figure 12: Results from strain-rate and temperature independent 3D single (shell) element (Q4GS ) tests(C = 0, m = 0 and DC 1). Numerical results in EPX compared to analytical solution in Eq. (41).

3D continuum element

Finally, the material model is verified for a 3D (hexahedron) continuum element (CUBE ) in uniaxial tensionand simple shear (see Figure 13). Again, note that the JAUM option must be activated when using CEAcontinuum elements (e.g. CUBE ) to treat large strains.

29

0 0.5 1 1.5 20

200

400

600

800

1000

1200

Plastic strain p [mm/mm]

Str

ess [

MP

a]

σeq

(EPX)

R(p) (EPX)

σeq

(analytical)

R(p) (analytical)

(a) Uniaxial tension (CUBE)

0 0.5 1 1.5 20

200

400

600

800

1000

1200

Plastic strain p [mm/mm]

Str

ess [

MP

a]

σeq

(EPX)

R(p) (EPX)

σeq

(analytical)

R(p) (analytical)

(b) Simple shear (CUBE)

Figure 13: Results from strain-rate and temperature independent 3D single element (CUBE ) tests (C = 0,m = 0 and DC 1). Numerical results in EPX compared to analytical solution in Eq. (41).

It is seen that the equivalent stress σeq from the numerical tests is in excellent agreement with the analyticalsolution in Eq. (41) for all stress states (Figures 10 - 13), and that the material model is capable of describingthe rate-independent behaviour during plastic straining. This is also confirmed by comparing the equivalentstress σeq and the hardening variable R(p) at the end of the simulations (p = 2.0) in Table 3. Note thatnegative values of ∆σeq and ∆R(p) imply that the analytical value is larger than that in the respectivenumerical test.

Table 3: Summary of single element tests for strain rate and temperature independent behaviour at the endof the simulations (p = 2.0).

Elementσeq ∆σeq R(p = 2.0) ∆R(p = 2.0)

[MPa] [%] [MPa] [%]Analytical 1014.5 - 644.5 -

FUN3 (uniaxial) 1015.1 0.06 645.1 0.10Q42L (uniaxial) 1014.1 -0.04 644.1 -0.06

Q42L (shear) 1013.9 -0.06 643.9 -0.09Q4GS (uniaxial) 1014.1 -0.04 644.1 -0.06

Q4GS (shear) 1013.9 -0.06 643.9 -0.09CUBE (uniaxial) 1014.9 0.04 644.9 0.06

CUBE (shear) 1014.0 -0.04 644.0 -0.07

6.1.2. Viscoplasticity

Similar tests are performed by also including viscoplasticity (C = 0.01) in the model. This will verify if themodel is capable of describing the viscoplastic response at elevated strain rates by varying the plastic strainrate p according to Table 2. That is, starting with a plastic strain rate p = 1 and increasing this successivelyby one order of magnitude until p = 1000. The constant strain rates are obtained in the numerical tests byusing Eq. (102) (or Eq. (103)). As for the rate-independent tests, the temperature sensitivity and elementerosion are excluded from the numerical tests by choosing m = 0 and DC 1. The analytical solution thenreduces to the two first terms in Eq. (41), against which the numerical results will be compared. Finally,the rate-independent response (p ≤ 5e−4) from Section 6.1.1 is also plotted for comparison.

30

1D bar element

Figure 14 shows the performance of the model in describing the rate-dependent behaviour in uniaxial tensionfor a bar element (FUN3 ).

0 0.5 1 1.5 20

200

400

600

800

1000

1200

Plastic strain p [mm/mm]

Str

ess [

MP

a]

σeq (p ≤ 5e−4, EPX)σeq (p = 1, EPX)σeq (p = 10, EPX)σeq (p = 100, EPX)σeq (p = 1000, EPX)σeq (analytical)

Figure 14: Results from temperature independent 1D single element (FUN3 ) test in uniaxial tension (m = 0and DC 1). Numerical results in EPX compared to the first two terms of the analytical solution inEq. (41).

2D continuum element

Figure 15 shows the capabilities of the model in describing the viscoplastic response at elevated strain ratesfor the Q42L element in uniaxial tension (Figure 15a) and simple shear (Figure 15b).

0 0.5 1 1.5 20

200

400

600

800

1000

1200

Plastic strain p [mm/mm]

Str

ess [

MP

a]

σeq (p ≤ 5e−4, EPX)σeq (p = 1, EPX)σeq (p = 10, EPX)σeq (p = 100, EPX)σeq (p = 1000, EPX)σeq (analytical)

(a) Uniaxial tension (Q42L)

0 0.5 1 1.5 20

200

400

600

800

1000

1200

Plastic strain p [mm/mm]

Str

ess [

MP

a]

σeq (p ≤ 5e−4, EPX)σeq (p = 1, EPX)σeq (p = 10, EPX)σeq (p = 100, EPX)σeq (p = 1000, EPX)σeq (analytical)

(b) Simple shear (Q42L)

Figure 15: Results from temperature independent 2D single element (Q42L) tests (m = 0 and DC 1).Numerical results in EPX compared to the first two terms of the analytical solution in Eq. (41).

3D shell element

The same tests are performed for the Q4GS element in uniaxial tension (Figure 16a) and simple shear

31

(Figure 16b), where the numerical results are compared to the first two terms of the analytical solution inEq. (41).

0 0.5 1 1.5 20

200

400

600

800

1000

1200

Plastic strain p [mm/mm]

Str

ess [

MP

a]

σeq (p ≤ 5e−4, EPX)σeq (p = 1, EPX)σeq (p = 10, EPX)σeq (p = 100, EPX)σeq (p = 1000, EPX)σeq (analytical)

(a) Uniaxial tension (Q4GS)

0 0.5 1 1.5 20

200

400

600

800

1000

1200

Plastic strain p [mm/mm]

Str

ess [

MP

a]

σeq (p ≤ 5e−4, EPX)σeq (p = 1, EPX)σeq (p = 10, EPX)σeq (p = 100, EPX)σeq (p = 1000, EPX)σeq (analytical)

(b) Simple shear (Q4GS)

Figure 16: Results from temperature independent 3D single (shell) element (Q4GS ) tests (m = 0 andDC 1). Numerical results in EPX compared to the first two terms of the analytical solution in Eq. (41).

3D continuum element

Finally, the viscoplastic response of the material model is verified for the CUBE element in uniaxial tension(Figure 17a) and simple shear (Figure 17b).

0 0.5 1 1.5 20

200

400

600

800

1000

1200

Plastic strain p [mm/mm]

Str

ess [

MP

a]

σeq (p ≤ 5e−4, EPX)σeq (p = 1, EPX)σeq (p = 10, EPX)σeq (p = 100, EPX)σeq (p = 1000, EPX)σeq (analytical)

(a) Uniaxial tension (CUBE)

0 0.5 1 1.5 20

200

400

600

800

1000

1200

Plastic strain p [mm/mm]

Str

ess [

MP

a]

σeq (p ≤ 5e−4, EPX)σeq (p = 1, EPX)σeq (p = 10, EPX)σeq (p = 100, EPX)σeq (p = 1000, EPX)σeq (analytical)

(b) Simple shear (CUBE)

Figure 17: Results from temperature independent 3D single (continuum) element (CUBE ) tests (m = 0 andDC 1). Numerical results in EPX compared to the first two terms of the analytical solution in Eq. (41).

It is seen that the equivalent stress σeq from the numerical tests is in excellent agreement with the firsttwo terms of the analytical solution in Eq. (41) for all stress states (Figures 14 - 17), and that the materialmodel is capable of describing the viscoplastic response at elevated strain rates. This is also confirmed bycomparing the equivalent stress σeq and the hardening variable R(p) at the end of the simulations (p = 2.0)

32

in Table 4. By comparing to the results in Table 3 it is seen that the hardening variable R(p) is onlydependent on the plastic straining and independent of strain rate (as expected). Again, note that negativevalues of ∆σeq and ∆R(p) imply that the analytical value is larger than that in the respective numericaltest. Also note that the analytical reference in Table 4 is found using the plastic strain and temperaturehistory from the uniaxial tests with the FUN3 element, while the values given for ∆σeq and ∆T (p = 2.0)for each element and numerical test are found using the corresponding histories from the respective tests.That is, the analytical reference used to compute ∆σeq and ∆T (p = 2.0) in Table 4 is found using the plasticstrain and temperature history from the respective test.

33

Table 4: Summary of single element tests for temperature independent behaviour at the end of the simula-tions (p = 2.0).

Plastic strain rateElement

σeq ∆σeq R(p = 2.0) ∆R(p = 2.0)[s−1] [MPa] [%] [MPa] [%]

≤ 5e−4

Analytical 1014.5 - 644.5 -FUN3 (uniaxial) 1015.1 0.06 645.1 0.10Q42L (uniaxial) 1014.1 -0.04 644.1 -0.06

Q42L (shear) 1013.9 -0.06 643.9 -0.09Q4GS (uniaxial) 1014.1 -0.04 644.1 -0.06

Q4GS (shear) 1013.9 -0.06 643.9 -0.09CUBE (uniaxial) 1014.9 0.04 644.9 0.06

CUBE (shear) 1014.0 -0.04 644.0 -0.07

1

Analytical 1094.6 - 644.4 -FUN3 (uniaxial) 1094.6 0.0 644.4 0.0Q42L (uniaxial) 1094.6 0.0 644.4 0.0

Q42L (shear) 1094.6 0.0 644.5 0.0Q4GS (uniaxial) 1094.6 0.0 644.4 0.0

Q4GS (shear) 1094.6 0.0 644.5 0.0CUBE (uniaxial) 1094.6 0.0 644.5 0.0

CUBE (shear) 1094.6 0.0 644.5 0.0

10

Analytical 1120.1 - 644.4 -FUN3 (uniaxial) 1120.1 0.0 644.4 0.0Q42L (uniaxial) 1120.1 0.0 644.4 0.0

Q42L (shear) 1120.1 0.0 644.5 0.0Q4GS (uniaxial) 1120.1 0.0 644.5 0.0

Q4GS (shear) 1120.1 0.0 644.5 0.0CUBE (uniaxial) 1120.1 0.0 644.5 0.0

CUBE (shear) 1120.1 0.0 644.5 0.0

100

Analytical 1146.1 - 644.4 -FUN3 (uniaxial) 1146.1 -0.01 644.4 -0.02Q42L (uniaxial) 1146.1 -0.01 644.4 -0.01

Q42L (shear) 1146.1 -0.01 644.4 -0.01Q4GS (uniaxial) 1146.1 -0.01 644.4 -0.01

Q4GS (shear) 1146.1 -0.01 644.4 -0.01CUBE (uniaxial) 1146.1 -0.01 644.4 -0.01

CUBE (shear) 1146.1 0.0 644.4 -0.01

1000

Analytical 1172.8 - 644.4 -FUN3 (uniaxial) 1171.9 -0.08 643.6 -0.13Q42L (uniaxial) 1172.0 -0.07 643.7 -0.11

Q42L (shear) 1172.2 -0.06 643.9 -0.09Q4GS (uniaxial) 1172.0 -0.07 643.7 -0.11

Q4GS (shear) 1172.2 -0.06 643.9 -0.09CUBE (uniaxial) 1172.2 -0.05 643.9 -0.08

CUBE (shear) 1172.4 -0.04 644.0 -0.07

34

6.1.3. Thermoviscoplasticity

Now, thermoviscoplasticity is also included in the model by specifying m = 1.0 in the material input (ac-cording to Table 1). The temperature T from EPX was used as input in Eq. (41) to obtain the analyticalsolution for the equivalent stress σeq, while the incremental plastic strain ∆p and equivalent stress σeq fromthe numerical solution is used as input in Eq. (45) to obtain an analytical solution for the temperatureevolution.

1D bar element

Figure 18 shows the performance of the model in describing the thermoviscoplastic behaviour (Figure 18a)and temperature evolution (Figure 18b) in uniaxial tension for a bar element (FUN3 ).

0 0.5 1 1.5 20

200

400

600

800

1000

1200

Plastic strain p [mm/mm]

Str

ess [

MP

a]

σeq (p ≤ 5e−4, EPX)σeq (p = 1, EPX)σeq (p = 10, EPX)σeq (p = 100, EPX)σeq (p = 1000, EPX)σeq (analytical)

(a) Equivalent stress σeq

0 0.5 1 1.5 20

100

200

300

400

500

600

700

800

Plastic strain p [mm/mm]

Tem

pera

ture

[K

]

T (p) (p ≤ 5e−4, EPX)T (p) (p = 1, EPX)T (p) (p = 10, EPX)T (p) (p = 100, EPX)T (p) (p = 1000, EPX)T (p) (analytical)

(b) Temperature T

Figure 18: Results from temperature dependent 1D single element (FUN3 ) test in uniaxial tension (m = 1.0and DC 1). Numerical results in EPX compared to the analytical solution in Eq. (41) and Eq. (45).

2D continuum element

Figure 19 shows the capabilities of the computational model in predicting the thermoviscoplastic responseat elevated strain rates for the Q42L element in uniaxial tension (Figure 19a) and simple shear (Figure 19b).

35

0 0.5 1 1.5 20

200

400

600

800

1000

1200

Plastic strain p [mm/mm]

Str

ess [

MP

a]

σeq (p ≤ 5e−4, EPX)σeq (p = 1, EPX)σeq (p = 10, EPX)σeq (p = 100, EPX)σeq (p = 1000, EPX)σeq (analytical)

(a) Equivalent stress σeq (uniaxial tension)

0 0.5 1 1.5 20

200

400

600

800

1000

1200

Plastic strain p [mm/mm]

Str

ess [

MP

a]

σeq (p ≤ 5e−4, EPX)σeq (p = 1, EPX)σeq (p = 10, EPX)σeq (p = 100, EPX)σeq (p = 1000, EPX)σeq (analytical)

(b) Equivalent stress σeq (simple shear)

0 0.5 1 1.5 20

100

200

300

400

500

600

700

800

Plastic strain p [mm/mm]

Tem

pera

ture

[K

]

T (p) (p ≤ 5e−4, EPX)T (p) (p = 1, EPX)T (p) (p = 10, EPX)T (p) (p = 100, EPX)T (p) (p = 1000, EPX)T (p) (analytical)

(c) Temperature T (uniaxial tension)

0 0.5 1 1.5 20

100

200

300

400

500

600

700

800

Plastic strain p [mm/mm]

Tem

pera

ture

[K

]

T (p) (p ≤ 5e−4, EPX)T (p) (p = 1, EPX)T (p) (p = 10, EPX)T (p) (p = 100, EPX)T (p) (p = 1000, EPX)T (p) (analytical)

(d) Temperature T (simple shear)

Figure 19: Results from temperature dependent (m = 1.0 and DC 1) 2D single (continuum) element(Q42L) tests in uniaxial tension (left) and simple shear (right). Numerical results in EPX compared to theanalytical solution in Eq. (41) and Eq. (45).

3D shell element

The same tests are performed for the Q4GS element in uniaxial tension (Figure 20a) and simple shear(Figure 20b).

36

0 0.5 1 1.5 20

200

400

600

800

1000

1200

Plastic strain p [mm/mm]

Str

ess [

MP

a]

σeq (p ≤ 5e−4, EPX)σeq (p = 1, EPX)σeq (p = 10, EPX)σeq (p = 100, EPX)σeq (p = 1000, EPX)σeq (analytical)

(a) Equivalent stress σeq (uniaxial tension)

0 0.5 1 1.5 20

200

400

600

800

1000

1200

Plastic strain p [mm/mm]

Str

ess [

MP

a]

σeq (p ≤ 5e−4, EPX)σeq (p = 1, EPX)σeq (p = 10, EPX)σeq (p = 100, EPX)σeq (p = 1000, EPX)σeq (analytical)

(b) Equivalent stress σeq (simple shear)

0 0.5 1 1.5 20

100

200

300

400

500

600

700

800

Plastic strain p [mm/mm]

Tem

pera

ture

[K

]

T (p) (p ≤ 5e−4, EPX)T (p) (p = 1, EPX)T (p) (p = 10, EPX)T (p) (p = 100, EPX)T (p) (p = 1000, EPX)T (p) (analytical)

(c) Temperature T (uniaxial tension)

0 0.5 1 1.5 20

100

200

300

400

500

600

700

800

Plastic strain p [mm/mm]

Tem

pera

ture

[K

]

T (p) (p ≤ 5e−4, EPX)T (p) (p = 1, EPX)T (p) (p = 10, EPX)T (p) (p = 100, EPX)T (p) (p = 1000, EPX)T (p) (analytical)

(d) Temperature T (simple shear)

Figure 20: Results from temperature dependent (m = 1.0 and DC 1) 3D single (shell) element (Q4GS )tests in uniaxial tension (left) and simple shear (right). Numerical results in EPX compared to the analyticalsolution in Eq. (41) and Eq. (45).

3D continuum element

Finally, the thermoviscoplastic response of the material model is verified for the CUBE element in uniaxialtension (Figure 21a) and simple shear (Figure 21b).

37

0 0.5 1 1.5 20

200

400

600

800

1000

1200

Plastic strain p [mm/mm]

Str

ess [

MP

a]

σeq (p ≤ 5e−4, EPX)σeq (p = 1, EPX)σeq (p = 10, EPX)σeq (p = 100, EPX)σeq (p = 1000, EPX)σeq (analytical)

(a) Equivalent stress σeq (uniaxial tension)

0 0.5 1 1.5 20

200

400

600

800

1000

1200

Plastic strain p [mm/mm]

Str

ess [

MP

a]

σeq (p ≤ 5e−4, EPX)σeq (p = 1, EPX)σeq (p = 10, EPX)σeq (p = 100, EPX)σeq (p = 1000, EPX)σeq (analytical)

(b) Equivalent stress σeq (simple shear)

0 0.5 1 1.5 20

100

200

300

400

500

600

700

800

Plastic strain p [mm/mm]

Tem

pera

ture

[K

]

T (p) (p ≤ 5e−4, EPX)T (p) (p = 1, EPX)T (p) (p = 10, EPX)T (p) (p = 100, EPX)T (p) (p = 1000, EPX)T (p) (analytical)

(c) Temperature T (uniaxial tension)

0 0.5 1 1.5 20

100

200

300

400

500

600

700

800

Plastic strain p [mm/mm]

Tem

pera

ture

[K

]

T (p) (p ≤ 5e−4, EPX)T (p) (p = 1, EPX)T (p) (p = 10, EPX)T (p) (p = 100, EPX)T (p) (p = 1000, EPX)T (p) (analytical)

(d) Temperature T (simple shear)

Figure 21: Results from temperature dependent (m = 1.0 and DC 1) 3D single (continuum) element(CUBE ) tests in uniaxial tension (left) and simple shear (right). Numerical results in EPX compared to theanalytical solution in Eq. (41) and Eq. (45).

Again, it is seen that the equivalent stress σeq from the numerical tests is in excellent agreement with theanalytical solution in Eq. (41) for all stress states (Figures 18 - 21), and that the computational model iscapable of describing the temperature softening of the equivalent stress due to adiabatic heating. This is alsosupported by comparing the equivalent stress σeq and the temperature T at the end of the simulations (p =2.0) in Table 5. Note that the analytical reference in Table 5 is found using the plastic strain and temperaturehistory from the uniaxial tests with the FUN3 element, while the values given for ∆σeq and ∆T (p = 2.0)for each element and numerical test are found using the corresponding histories from the respective tests.That is, the analytical reference used to compute ∆σeq and ∆T (p = 2.0) in Table 5 is found using the plasticstrain and temperature history from the respective test. Also note that in a thermoviscoplastic model thereis always a competition between viscoplasticity and thermal softening, where (viscoplastic) strain hardeningimplies an increase of strength locally in the material and distributes the plasticity while thermal softeningresults in localization of plasticity.

38

Table 5: Summary of single element tests for thermo-viscoplastic behaviour behaviour at the end of thesimulations (p = 2.0).

Plastic strain rateElement

σeq ∆σeq T (p = 2.0) ∆T (p = 2.0)[s−1] [MPa] [%] [K] [%]

≤ 5e−4

Analytical 733.0 - 711.3 -FUN3 (uniaxial) 733.7 0.10 711.2 -0.02Q42L (uniaxial) 732.9 -0.07 710.5 -0.01

Q42L (shear) 732.6 -0.05 711.1 -0.01Q4GS (uniaxial) 732.9 -0.07 710.5 -0.01

Q4GS (shear) 732.6 -0.05 711.1 -0.01CUBE (uniaxial) 733.5 0.06 710.9 -0.01

CUBE (shear) 732.7 -0.04 711.1 -0.01

1

Analytical 771.1 - 738.4 -FUN3 (uniaxial) 771.1 0.0 738.3 -0.02Q42L (uniaxial) 771.1 0.0 738.3 -0.02

Q42L (shear) 770.7 0.0 738.9 -0.02Q4GS (uniaxial) 771.1 0.0 738.2 -0.02

Q4GS (shear) 770.7 0.0 738.9 -0.02CUBE (uniaxial) 771.1 0.0 738.3 -0.02

CUBE (shear) 770.7 0.0 738.9 -0.02

10

Analytical 782.7 - 747.0 -FUN3 (uniaxial) 782.7 0.0 746.9 -0.01Q42L (uniaxial) 782.7 0.0 746.9 -0.01

Q42L (shear) 782.3 0.0 747.4 -0.02Q4GS (uniaxial) 782.7 0.0 746.9 -0.01

Q4GS (shear) 782.3 0.0 747.4 -0.02CUBE (uniaxial) 782.7 0.0 746.9 -0.01

CUBE (shear) 782.3 0.0 747.5 -0.02

100

Analytical 794.3 - 755.8 -FUN3 (uniaxial) 794.2 -0.01 755.6 -0.01Q42L (uniaxial) 794.2 -0.01 755.6 -0.01

Q42L (shear) 793.8 -0.01 756.3 -0.02Q4GS (uniaxial) 794.3 0.0 755.6 -0.01

Q4GS (shear) 793.8 -0.01 756.3 -0.02CUBE (uniaxial) 794.3 0.0 755.6 -0.01

CUBE (shear) 793.8 0.0 756.3 -0.02

1000

Analytical 806.0 - 764.5 -FUN3 (uniaxial) 805.5 -0.06 764.3 -0.02Q42L (uniaxial) 805.5 -0.07 764.3 -0.01

Q42L (shear) 805.2 -0.05 764.9 -0.01Q4GS (uniaxial) 805.6 -0.07 764.2 -0.01

Q4GS (shear) 805.2 -0.05 764.9 -0.01CUBE (uniaxial) 805.6 -0.05 764.3 -0.01

CUBE (shear) 805.2 -0.04 764.9 -0.01

39

6.1.4. Ductile failure

Finally, ductile failure is included in the model by specifying Dc = 1 in the material input (see Table 1).To couple the model with the element erosion option available in EPX, it is also necessary to include EROSin the input file under the directive defining the type of computation (typically CPLA LAGR EROS 1.0 orTRID LAGR EROS 1.0 ) [7]. Then, when the damage variable in Eq. (46) reaches its critical value D = DC ,the material (or integration) point stiffness is deleted and the element is removed. Note that the respectivematerial (or integration) point may fail either due to the CL criterion when Dc = 1.0 or when the materialmelts at T = Tm. The latter scenario is not included in these single element tests since the material meltsat very large strains for the material used in this study. However, the user should be aware that this typeof material failure may occur in problems with sufficiently large strains.

1D bar element

Figure 22 shows the performance of the VPJC model in predicting ductile failure during the rate-dependentbehaviour in uniaxial tension for a bar element (FUN3 ).

0 0.1 0.2 0.3 0.4 0.5 0.6 0.70

200

400

600

800

1000

1200

Plastic strain p [mm/mm]

Str

ess [M

Pa]

σeq (p ≤ 5e−4, EPX)

σeq (p = 1, EPX)

σeq (p = 10, EPX)

σeq (p = 100, EPX)

σeq (p = 1000, EPX)

Failure

Figure 22: Results from temperature and strain-rate dependent 1D single element (FUN3 ) test in uniaxialtension including failure (m = 1.0 and DC = 1).

2D continuum element

Figure 23 shows the capabilities of the model in predicting ductile failure during the rate-dependent behaviourat elevated strain rates for the Q42L element in uniaxial tension (Figure 23a) and simple shear (Figure 23b).

40

0 0.1 0.2 0.3 0.4 0.5 0.6 0.70

200

400

600

800

1000

1200

Plastic strain p [mm/mm]

Str

ess [M

Pa]

σeq (p ≤ 5e−4, EPX)

σeq (p = 1, EPX)

σeq (p = 10, EPX)

σeq (p = 100, EPX)

σeq (p = 1000, EPX)

Failure

(a) Uniaxial tension (Q42L)

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.10

200

400

600

800

1000

1200

Plastic strain p [mm/mm]

Str

ess [M

Pa]

σeq (p ≤ 5e−4, EPX)

σeq (p = 1, EPX)

σeq (p = 10, EPX)

σeq (p = 100, EPX)

σeq (p = 1000, EPX)

Failure

(b) Simple shear (Q42L)

Figure 23: Results from temperature and strain-rate dependent 2D single element (Q42L) tests includingfailure (m = 1.0 and DC = 1).

3D shell element

The same tests are performed for the Q4GS element in uniaxial tension (Figure 24a) and simple shear(Figure 24b).

0 0.1 0.2 0.3 0.4 0.5 0.6 0.70

200

400

600

800

1000

1200

Plastic strain p [mm/mm]

Str

ess [M

Pa]

σeq (p ≤ 5e−4, EPX)

σeq (p = 1, EPX)

σeq (p = 10, EPX)

σeq (p = 100, EPX)

σeq (p = 1000, EPX)

Failure

(a) Uniaxial tension (Q4GS)

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.10

200

400

600

800

1000

1200

Plastic strain p [mm/mm]

Str

ess [M

Pa]

σeq (p ≤ 5e−4, EPX)

σeq (p = 1, EPX)

σeq (p = 10, EPX)

σeq (p = 100, EPX)

σeq (p = 1000, EPX)

Failure

(b) Simple shear (Q4GS)

Figure 24: Results from temperature and strain-rate dependent 3D single (shell) element (Q4GS ) testsincluding failure (m = 1.0 and DC = 1).

3D continuum element

Finally, the models capability in predicting ductile failure is verified for the CUBE element in uniaxialtension (Figure 25a) and simple shear (Figure 25b).

41

0 0.1 0.2 0.3 0.4 0.5 0.6 0.70

200

400

600

800

1000

1200

Plastic strain p [mm/mm]

Str

ess [M

Pa]

σeq (p ≤ 5e−4, EPX)

σeq (p = 1, EPX)

σeq (p = 10, EPX)

σeq (p = 100, EPX)

σeq (p = 1000, EPX)

Failure

(a) Uniaxial tension (CUBE)

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.10

200

400

600

800

1000

1200

Plastic strain p [mm/mm]

Str

ess [M

Pa]

σeq (p ≤ 5e−4, EPX)

σeq (p = 1, EPX)

σeq (p = 10, EPX)

σeq (p = 100, EPX)

σeq (p = 1000, EPX)

Failure

(b) Simple shear (CUBE)

Figure 25: Results from temperature and strain-rate dependent 3D single (continuum) element (CUBE )tests including failure (m = 1.0 and DC = 1).

Figures 22 - 25 show that the VPJC material model accounts for viscoplasticity when the CL criterion isused to predict ductile failure. That is, elevated strain rates result in reduced elongation to failure due tothe viscoplastic increase in the equivalent stress. This is explained by Eq. (20) where an increased strainrate results in an increased flow stress σy(σ0, p, p). Consequently, the equivalent stress will increase to fulfillthe yield criterion (fv(σ) = 0). This also results in an increase in the major principal stress σ1 and that thedamage parameter Wc in Eq. (46) is reached at smaller elongations. The difference in elongation at failurepf in uniaxial tension and simple shear is mainly explained by the dependence of stress triaxiality σ∗ andLode parameter µσ in Eqs. (46)-(51). Note that the differences observed in failure strain in Figures 23 - 25corresponds well with the corresponding difference predicted by Eq. (51) in Figure 6.

6.2. Tension test

The material parameters (Q1, C1, Q2 and C2) in the first multiplicative term in Eq. (41) and CL parameterWc in Eq. (46) were determined based on inverse modelling of the quasi-static tension tests presented in[21]. These tension tests therefore serve as an excellent basis for validation of the quasi-static part of theVPJC material model.

The uniaxial tension tests were modelled based on the experimental geometry (see Figure 26a) and pre-scribing the same boundary conditions as in the material tests (see Figure 26b). The numerical model wasdiscretized in 3116 4-node quadrilaterals where the initial element size in the gauge area was equal to thethickness (0.8 mm) in an attempt to capture the diffuse and local necking observed in the experiments.Two simulations were performed using both the Q42L and Q4GS elements, where the former is a planestress elements with 2 dofs per node and 4 integration points (for full integration) while the latter is a 3Dshell element with 6 dofs per node and 20 integration points (5 through the thickness) [7]. Mass-scalingby increasing the density by a factor 109 (ρ = 7850E9) was used to speed up the simulation, because anexplicit solution in time results in significant computational times (i.e., a considerable amount of time in-crements) for quasi-static tests. The quasi-static nature of the test was ensured by comparing the kineticenergy present in the simulations to the plastic dissipation and found negligible. Moreover, since the plasticresponse of the material cannot be assumed to occur within a short time duration in a quasi-static test, theassumption of an adiabatic condition is no longer valid and the simulation should be run without thermalsoftening (m = 0). It was also assumed negligible viscoplasticity in these tests (C = 0), while the remainingmaterial parameters and physical constants were taken from Table 1. The reader is referred to [21, 27] and

42

[29] for more information about the experimental work and numerical model, respectively, for these tensiontests.

70

200

12.5

40 φ12.7R15

25.05

50.05

(a) Geometry (experimental)

y

x vx = 2.1mm/min

Nodal constraints

Nodal loads

70

(b) Numerical model

Figure 26: Geometry and numerical model of the uniaxial tension test.

A comparison of the experimental and numerical results in terms of nominal stress-strain curves is presentedin Figure 27. It is seen that the numerical simulations are in excellent agreement with the experimental data,and the quasi-static behaviour of the VPJC material model may be assumed to be properly implementedin EPX.

0 0.05 0.1 0.15 0.2 0.25 0.30

100

200

300

400

500

600

700

800

Nominal strain [mm/mm]

No

min

al str

ess [

MP

a]

Experiment 0°

Experiment 45°

Experiment 90°

FEA (Q42L)

FEA (Q4GS)W

c = 473 MPa

Wc = 473 MPa

Figure 27: Nominal stress-strain curves from uniaxial tension tests along three different loading directionsfor the Docol 600 DL steel. Numerical results from EPX (FEA) for both Q42L and Q4GS elements arecompared to the experimental data in [21]. The red and green dots denote the point of failure in thedetermination of Wc.

43

6.3. Airblast experiments

A series of airblast experiments on the same steel (Docol 600 DL) material were further used as an validationof the dynamic part of the VPJC material model. Note that the material tests by Holmen et al. [21] weretaken from the same plates (i.e., the same batch) as those used in this study. Thus, the material dataprovided in Table 1 also apply for these airblast experiments.

The experimental setup is shown in Figure 28, and consists of a steel mounting frame fixed to the concretefloor with outer dimensions 1.0 × 1.0 × 0.015 m3 and a square opening of 0.3 × 0.3 m2 at the centre. Thesquare plate specimen with dimensions of 0.4 × 0.4 × 0.0008 m3 were clamped to the rigid mounting frameusing M12 bolts and a clamping frame in an attempt to achieve fixed boundary conditions. The explosivemass W was positioned at stand-off distances R = 0.125 m, 0.250 m and 0.375 m relative to the centrepoint of the plate. The explosive material was Composition C-4 with a spherical charge, a mass of 30 g(equivalent to 40.2 g of TNT) and a diameter of approximately 34.5 mm (Figure ??). Finally, the plateswere painted with a speckle pattern (Figure 28c) and three-dimensional digital image correlation (3D-DIC)analyses were conducted to measure the transient displacement field during the tests. The stereovision setupof the high-speed cameras (Phantom v1610) is illustrated in Figure 28a. The reader is referred to [27] forfurther details regarding the experimental work.

Steel mounting frame

Blast pencil∼ 25

Test specimen

Clamping frame

Charge of 30 g C4

∼ 3300mmStand-off-distance

Cam 1

Cam 2

(a) Sketch of setup

(b) Spherical charge (c) DIC speckle pattern

Figure 28: Experimental setup airblast test [27]. The speckle pattern in (c) is seen from the cameras.

44

Figure 29 shows the assembly of the numerical model, where the symmetry of the problem was utilized tomodel only one quarter of the experimental setup using symmetric boundary conditions. The plate wasmodelled using a Lagrangian discretization with an element size of approximately 2.5 mm (Figure 29b)and Q4GS shell elements, while the bolts and clamping frames were represented by 8-node brick elements(CUB8 ) with 8 integration points and the VPJC model with a high elastic limit to ensure elastic behaviourusing the physical constants for steel in Table 1. The steel mounting frame and the bolts were modelled asone component (Figure 29a) and the bolts were modelled as stress-free. The pre-tensioning of the bolts wasaccounted for by applying an external pressure at the contact area (Ac = 175 mm2) between the bolt headsand the clamping frame (see Figure 29c). Dividing the pre-tensioning force Fp in each bolt by this contactarea resulted in a contact pressure of 527 MPa between each bolt head and the clamping frame. Contactbetween the plate, bolts and frames was modelled using a surface-to-surface contact algorithm (GLIS ). Thiscontact algorithm is based on slave nodes and master surfaces where contact was enforced by Lagrangianmultipliers when a slave node penetrated a master surface. The plate was modelled as the slave and thestatic friction coefficient between the plate and clamping frames was set to 0.15, while the dynamic frictioncoefficient was equal to 0.10. The blast loading was applied on the plate by using the AIRB last directive inEPX where the mass and positioning of the charge were used as input [7]. A more detailed description ofthe numerical model is given in [29].

(a) Steel mounting frame and bolts (b) Plate specimen (c) Complete clamping assembly

Figure 29: Numerical model of the airblast experiments showing (a) steel mounting frame and bolts as onecomponent (in cyan), (b) plate specimen (in green) and (c) complete assembly including the clamping frame(also in cyan) and contact area between bolt heads and clamping frame (in magenta) used to model theeffect of the pre-tensioning of the bolts [29].

A comparison of the experimental and numerical results in terms of mid-point deflection versus time is pre-sented in Figure 30. Considering the complexity in these airblast tests, it is observed a very good agreementbetween the numerical simulations and experimental observations. In particular, the maximum displacementcorresponds very well with the experimental data for the two largest stand-off distances and the reversedsnap buckling is captured at the largest stand-off distance. The representation of the maximum displacementis also quite good at the closest stand-off distance, however, the displacement is somewhat underestimated.This is probably due to poorer representation of the actual loading on the plate, since we are outside therange of the empirical methods used in the AIRB directive (Z = R/W 1/3 < 0.4 m/kg1/3). However, sincethe scope of this report is to validate the implementation of the VPJC model this is not given any moreconsideration. Thus, the VPJC material model is found to perform excellent also in fast transient dynamics.

45

0 2 4 6 8 10

−15

−10

−5

0

5

10

15

20

25

30

35

Time [ms]

Mid

−p

oin

t d

efle

ctio

n [

mm

]

Experiments

FEA (R = 0.125 m)

FEA (R = 0.250 m)

FEA (R = 0.375 m)

Figure 30: Comparison of experimental and numerical results (FEA) for the airblast tests [29].

7. Concluding Remarks

An elastic-thermoviscoplastic Johnson-Cook type constitutive model was implemented in EUROPLEXUS(EPX) and called VPJC. The VPJC model allows the user to choose between either a fully vectorized ver-sion of the forward Euler integration (i.e., cutting plane) algorithm or a special case of the backward Euler(i.e., radial return) algorithm. Ductile failure is also included in the material model through the criterionproposed by Cockcroft and Latham. This allows the model to be coupled with the element erosion optionavailable in EPX. The material model is applicable for applicable for one-, two- and three-dimensional stressstates and a wide range of elements in EPX.

The performance of the model was first verified using single element simulations in uniaxial tension andsimple shear. The model was found to respond very well to all the relevant test cases in terms of rate-independent, viscoplastic and thermoviscoplastic behaviour. The dependence of the stress triaxiality on thefailure strain was also shown by including ductile failure in the single element verification. Finally, to ensurea realistic behaviour, the VPJC model was validated against experimental data from quasi-static tensiontests and airblast experiments. It was found that the numerical simulations were in excellent agreementwith the experimental data, and the VPJC model is therefore found to be properly implemented in EPX.

8. Further work

1. The hardening (Power law) given in Eq. (40) could be included in the model. By introducing a FLAGin the material card the user could then choose between Voce hardening (e.g. FLAG = 0) or Powerlaw hardening (e.g. FLAG = 1). This may be done in the plastic corrector by

IF FLAG = 0 THEN use Voce hardening from Eq. (42)

HR = Q1C1 exp(−C1p) +Q2C2 exp(−C2p)

ELSEIF FLAG = 1 THEN use Power law hardening from Eq. (40)

HR = Bnpn−1

ELSE Material routine stops and print message:

’VPJC - INVALID TYPE OF HARDENING, FLAG=0 or 1’

It should be noted that the material card must be extended to include three more parameters, i.e., B,n and a FLAG.

46

2. Extend the radial return method to include plane stress. This could be done by constraining theoriginal three-dimensional constitutive equation to obtain a plane stress state. That is, prescribe onlythe in-plane strains in the three-dimensional equation and use plane stress constraints as an additionalcondition to determine the in-plane stresses and out-of-plane strains. The reader is referred to [5](Chapter 9) and [26] for more information on this approach.

3. A substepping scheme could be implemented to increase the stability and accuracy of the return mapalgorithms.

4. Implement and add an option of using the Johnson-Cook failure criterion [30]. This is done by addingthe following during the update of variables depending on DLAMBDA

C COMPUTE FAILURE STRAIN BASED ON JOHNSON-COOK

PF = (D1 +D2 ∗ EXP (−D3 ∗ SIGMAH/PHI))

. ∗ ((1 +DLAMBDA/(PDOT ∗DT )) ∗ ∗D4)

. ∗ (1 +D5 ∗ ((T − TR)/(TM − TR)))

C COMPUTE AND UPDATE DAMAGE VARIABLE

DJC = DJC +DLAMBDA/PF

∗...continue as is ...

∗990 IF (D >= DC OR DJC >= DC) THEN

∗∗ the current gp was already failed or is failing now ...

...continue as is ...

IF (D >= DC) THEN

WRITE(59,*) ’Element deleted due to CL-criterion’

IF (DJC >= DC) THEN

WRITE(59,*) ’Element deleted due to JC-criterion’

...continue as is ...

Note that the material input must be extended to include five more parameters, i.e., D1, D2, D3, D4

and D5. Material constants for a Weldox steel can be found in [14], i.e., D1 = 0.0705, D2 = 1.732,D3 = −0.54, D4 = −0.015 and D5 = 0.

References

[1] T. Belytschko, W. Liu, B. Moran, Nonlinear Finite Elements for Continua and Structures, John Wiley & Sons (2000).[2] M. Ortiz, J. Simo, An analysis of a new class of integration algorithms for elastoplastic relations, International Journal of

Numerical Methods in Engineering 23 (1986) 353–366.[3] J. Simo, R. L. Taylor, Consistent tangent operators for rate-independent elastoplasticity, Computer Methods in Applied

Mechanics and Engineering 48 (3) (1985) 101–118.[4] R. D. Krieg, D. B. Krieg, Accuracies of numerical solution methods for elastic-perfectly plastic model, ASME Journal of

Pressure Vessel Technology 99 (4) (1977) 510–515.[5] E. A. de Souza Neto, D. Peric, D. R. J. Owen, Computational methods for plasticity : theory and applications, Wiley,

UK (2008).[6] G. R. Johnson, W. H. Cook, A Constitutive Model and Data for Metals Subjected to Large Strains, High Strain Rates

and High Temperatures, Proceedings of the 7th International Symposium on Ballistics, Hague (1983) 541–547.[7] Europlexus, User’s manual, Joint Research Centre, http://europlexus.jrc.ec.europa.eu/public/manual html/index.html

[cited 07.06.16].

47

[8] S. A. Khan, S. Huang, Continuum Theory of Plasticity, Wiley-Interscience, 1995.[9] M. G. Cockcroft, D. J. Latham, Ductility and the workability of metals, Journal of the institute of metals 96 (1968) 33–39.

[10] O. S. Hopperstad, T. Børvik, Lecture Notes KT8306 Plasticity Theory, Department of Structural Engineering, SIMLab(2013).

[11] J. Lubliner, Plasticity theory, Dover (2008).[12] F. Irgens, Continuum Mechanics, Springer-Verlag (2008).[13] M. Ortiz, G. T. Camacho, Adaptive Lagrangian modelling of ballistic penetration metallic targets, Comput. Methods

Appl. Mech. Engrg. 142 (1997) 269–301.[14] T. Børvik, O. S. Hopperstad, T. Berstad, M. Langseth, A computational model of viscoplasticity and ductile damage for

impact and penetration, European Journal of Mechanics - A/Solids 20 (2001) 685–712.[15] E. Voce, The relationship between stress and strain for homogenous deformation, Journal of the Institute of Metals 74

(1948) 536–562.[16] T. L. Anderson, Fracture Mechanics - Fundamentals and Applications, Taylor and Francis Group, third edition (2005).[17] J. Lemaitre, J.-L. Chabocke, Mechanics of Solid Materials, Cambridge University Press.[18] T. Børvik, S. Dey, A. H. Clausen, Perforation resistance of five different high-strength plates subjected to small-arms

projectiles, International Journal of Impact Engineering 36 (2009) 948–964.[19] A. Kane, T. Børvik, A. Benallal, O. S. Hopperstad, Failure criteria with unilateral conditions for simulation of plate

perforation, European Journal of Mechanics - A/Solids 30 (2011) 468–476.[20] S. Dey, T. Børvik, O. S. Hopperstad, M. Langseth, On the influence of fracture criterion in projectile impact of steel

plates, Computational Materials Science 38 (2006) 176–191.[21] J. K. Holmen, O. S. Hopperstad, T. Børvik, Low-velocity impact on multi-layered dual-phase steel plates, International

Journal of Impact Engineering 78 (2015) 161–177.[22] J. K. Holmen, J. Johnsen, O. S. Hopperstad, T. Børvik, Influence of fragmentation on the capacity of aluminium alloy

plates subjected to ballistic impact, European Journal of Mechanics A/Solids 55 (2016) 221–233.[23] M. L. Wilkins, Calculation of elastic-plastic flow, Technical report UCRL-7322, Lawrence radiation Laboratory, University

of California. Livermore, California April 19.[24] R. P. R. Cardoso, J. W. Yoon, Stress integration method for a nonlinear kinematic/isotropic hardening model and its

characterization based on polycrystal plasticity, International Journal of Plasticity 25 (2009) 1684–1710.[25] R. de Borst, P. H. Feenstra, Studies in anisotropic plasticity with reference to the Hill criterion, International Journal for

Numerical Methods in Engineering 29 (1990) 315–336.[26] J. C. Simo, R. L. Taylor, A Return Mapping Algorithm for Plane Stress Elastoplasticity, International Journal for Nu-

merical Methods in Engineering 22 (1986) 649–670.[27] V. Aune, E. Fagerholt, K. Hauge, M. Langseth, T. Børvik, Experimental study on the response of aluminium and

steel plates subjected to free airblast loading, International Journal of Impact Engineering 90 (2016) 106–121. doi:

10.1016/j.ijimpeng.2015.11.017.[28] F. Casadei, V. Aune, G. Valsamos, M. Larcher, Testing of the Johnson-Cook material model VPJC in EUROPLEXUS,

European Commission, Joint Research Centre, Technical Report, JRC 98848 EUR 27594 (OPOCE LB-NA-27594-EN-N)(2015) 1–62. doi:10.2788/02760.

[29] V. Aune, G. Valsamos, F. Casadei, M. Larcher, M. Langseth, T. Børvik, Numerical study on the structural response ofblast-loaded thin aluminium and steel plates, Submitted for journal publication (2016).

[30] G. R. Johnson, W. H. Cook, Fracture characteristics of three metals subjected to various strains, strain rates, temperaturesand pressures, Engineering Fracture Mechanics 21 (I) (1985) 31–48.

48

Appendix A. Cutting plane algorithm by using a linear Taylor-expansion

A linear Taylor-expansion around fn+1(σ,X) = fn+1(σ, σy, υp(p),Γp(Tn)) reads

f i+1n+1 = f in+1 +

∂f

∂σ

∣∣∣∣in+1

∆σn+1 +∂f

∂X

∣∣∣∣in+1

∆Xn+1 (A.1)

which may be written in terms of the internal variables X as

f i+1n+1 = f in+1 +

∂f

∂σn+1δσn+1 +

∂f

∂σy,n+1δσy,n+1 +

∂f

∂υp,n+1δυp,n+1 = 0 (A.2)

where the flow stress σy,n+1 and the following expressions are chosen as the internal variables X in theTaylor-expansion

υp(∆λn+1) =

(1 +

∆λn+1

p0∆tn+1

)CΓp(Tn) =

(1−

[Tn − TrTm − Tr

]m)(A.3)

Introducing the partial derivatives of the internal variables Eq. (A.2) reads

f i+1n+1 = f in+1 +

∂f

∂σn+1δσn+1︸ ︷︷ ︸

−C ∂f∂σn+1

δλi+1n+1

+∂f

∂σy,n+1︸ ︷︷ ︸(−1)υp,n+1Γp,n

δσy,n+1︸ ︷︷ ︸∂σy,n+1∂λn+1

δλi+1n+1

+∂f

∂υp,n+1︸ ︷︷ ︸(−1)σy,n+1Γp,n

δυp,n+1︸ ︷︷ ︸∂υp,n+1∂pn+1

δpi+1n+1

= 0 (A.4)

f i+1n+1 = f in+1 −

∂f

∂σn+1C

∂f

∂σn+1δλi+1

n+1 − υp,n+1Γp,n∂σy,n+1

∂λn+1δλi+1

n+1 − σy,n+1Γp,n∂υp,n+1

∂λn+1

δλi+1n+1 = 0 (A.5)

f i+1n+1 = f in+1 −

∂f

∂σn+1C

∂f

∂σn+1δλi+1

n+1 − υp,n+1Γp,nHR,n+1δλi+1n+1 − σy,n+1Γp,n

∂υp,n+1

∂λn+1

δλi+1n+1 = 0 (A.6)

Moreover, using p = λ = ∆λ∆t the internal variable controlling the viscoplasticity υp(p) may be expressed as

υp(∆λ) and Eq. (A.6) may be written as

f i+1n+1 = f in+1 −

∂f

∂σn+1C

∂f

∂σn+1δλi+1

n+1 − υp,n+1Γp,nHR,n+1δλi+1n+1 − σy,n+1Γp,n

∂υp,n+1

∂∆λn+1δλi+1

n+1 = 0 (A.7)

Finally, the incremental change δλ in the incremental plastic multiplier ∆λ is then expressed as

δλi+1n+1 =

f in+1

∂f∂σn+1

C ∂f∂σn+1

+ υp,n+1Γp,nHR,n+1 + σy,n+1Γp,n∂υp,n+1

∂∆λn+1

(A.8)

where

∂υp(∆λn+1)

∂∆λ=

C

p0∆t

(1 +

∆λn+1

p0∆t

)C−1

(A.9)

HR,n+1 =dRn+1

dpn+1=

2∑k=1

QkCke(−Ckλn+1) (A.10)

∂f

∂σ

T[C]

∂f

∂σ

(A.11)

49

[C] =

C1111 C1122 C1133 C1112 C1121 C1123 C1132 C1113 C1131

C2211 C2222 C2233 C2212 C2221 C2223 C2232 C2213 C2231

C3311 C3322 C3333 C3312 C3321 C3323 C3332 C3313 C3331

C1211 C1222 C1233 C1212 C1221 C1223 C1232 C1213 C1231

C2111 C2122 C2133 C2112 C2121 C2123 C2132 C2113 C2131

C2311 C2322 C2333 C2312 C2321 C2323 C2332 C2313 C2331

C3211 C3222 C3233 C3212 C3221 C3223 C3232 C3213 C3231

C1311 C1322 C1333 C1312 C1321 C1323 C1332 C1313 C1331

C3111 C3122 C3133 C3112 C3121 C3123 C3132 C3113 C3131

(A.12)

∂f

∂σ

=

∂f∂σ11∂f∂σ22∂f∂σ33∂f∂σ12∂f∂σ21∂f∂σ23∂f∂σ32∂f∂σ13∂f∂σ31

=

f,11

f,22

f,33

f,12

f,21

f,23

f,32

f,13

f,31

(A.13)

∂f

∂σ

T[C] =

f,11C1111 + f,22C2211 + f,33C3311 + f,12C1211 + f,21C2111 + f,23C2311 + f,32C3211 + f,13C1311 + f,31C3111

f,11C1122 + f,22C2222 + f,33C3322 + f,12C1222 + f,21C2122 + f,23C2322 + f,32C3222 + f,13C1322 + f,31C3122

f,11C1133 + f,22C2233 + f,33C3333 + f,12C1233 + f,21C2133 + f,23C2333 + f,32C3233 + f,13C1333 + f,31C3133

f,11C1112 + f,22C2212 + f,33C3312 + f,12C1212 + f,21C2112 + f,23C2312 + f,32C3212 + f,13C1312 + f,31C3112

f,11C1121 + f,22C2221 + f,33C3321 + f,12C1221 + f,21C2121 + f,23C2321 + f,32C3221 + f,13C1321 + f,31C3121

f,11C1123 + f,22C2223 + f,33C3323 + f,12C1223 + f,21C2123 + f,23C2323 + f,32C3223 + f,13C1323 + f,31C3123

f,11C1132 + f,22C2232 + f,33C3332 + f,12C1232 + f,21C2132 + f,23C2332 + f,32C3232 + f,13C1332 + f,31C3132

f,11C1113 + f,22C2213 + f,33C3313 + f,12C1213 + f,21C2113 + f,23C2313 + f,32C3213 + f,13C1313 + f,31C3113

f,11C1131 + f,22C2231 + f,33C3331 + f,12C1231 + f,21C2131 + f,23C2331 + f,32C3231 + f,13C1331 + f,31C3131

(A.14)

∂f

∂σ

T[C]

∂f

∂σ

=

(f,11C1111 + f,22C2211 + f,33C3311 + f,12C1211 + f,21C2111 + f,23C2311 + f,32C3211 + f,13C1311 + f,31C3111)f,11

+ (f,11C1122 + f,22C2222 + f,33C3322 + f,12C1222 + f,21C2122 + f,23C2322 + f,32C3222 + f,13C1322 + f,31C3122)f,22

+ (f,11C1133 + f,22C2233 + f,33C3333 + f,12C1233 + f,21C2133 + f,23C2333 + f,32C3233 + f,13C1333 + f,31C3133)f,33

+ (f,11C1112 + f,22C2212 + f,33C3312 + f,12C1212 + f,21C2112 + f,23C2312 + f,32C3212 + f,13C1312 + f,31C3112)f,12

+ (f,11C1121 + f,22C2221 + f,33C3321 + f,12C1221 + f,21C2121 + f,23C2321 + f,32C3221 + f,13C1321 + f,31C3121)f,21

+ (f,11C1123 + f,22C2223 + f,33C3323 + f,12C1223 + f,21C2123 + f,23C2323 + f,32C3223 + f,13C1323 + f,31C3123)f,23

+ (f,11C1132 + f,22C2232 + f,33C3332 + f,12C1232 + f,21C2132 + f,23C2332 + f,32C3232 + f,13C1332 + f,31C3132)f,32

+ (f,11C1113 + f,22C2213 + f,33C3313 + f,12C1213 + f,21C2113 + f,23C2313 + f,32C3213 + f,13C1313 + f,31C3113)f,13

+ (f,11C1131 + f,22C2231 + f,33C3331 + f,12C1231 + f,21C2131 + f,23C2331 + f,32C3231 + f,13C1331 + f,31C3131)f,13

(A.15)

[C] =

C1111 C1122 C1133 0 0 0 0 0 0C2211 C2222 C2233 0 0 0 0 0 0C3311 C3322 C3333 0 0 0 0 0 0

0 0 0 C1212 C1221 0 0 0 00 0 0 C2112 C2121 0 0 0 00 0 0 0 0 C2323 C2332 0 00 0 0 0 0 C3223 C3232 0 00 0 0 0 0 0 0 C1313 C1331

0 0 0 0 0 0 0 C3113 C3131

(A.16)

50

Due to the symmetry of the stress, σij = σji, it is seen that also

Cijkl = Cjikl (A.17)

Since the strain tensor also is symmetric, εkl = εlk, it is also assumed that

Cijkl = Cijlk (A.18)

and the 9x9 matrix in Eq. (A.16) may be reduced to the following 6x6 matrix

[C] =

C1111 C1122 C1133 0 0 0C2211 C2222 C2233 0 0 0C3311 C3322 C3333 0 0 0

0 0 0 C1212 0 00 0 0 0 C2323 00 0 0 0 0 C1313

(A.19)

Adding major symmetry, i.e. Cijkl = Cklij , to Eq. (A.19) and assuming an isotropic material where all axesare symmetry axes, i.e. c1 = C1111 = C2222 = C3333, c4 = C1122 = C2211 = C1133 = C3311 = C2233 = C3322

and c7 = C1212 = C2323 = C1313. Then, Eq. (A.15) reduces to

∂f

∂σ

T[C]

∂f

∂σ

=

(f,11C1111 + f,22C2211 + f,33C3311)f,11 + (f,11C1122 + f,22C2222 + f,33C3322)f,22

+ (f,11C1133 + f,22C2233 + f,33C3333)f,33

+ (f,12C1212 + f,21C2112)f,12 + (f,12C1221 + f,21C2121)f,21

+ (f,23C2323 + f,32C3223)f,23 + (f,23C2332 + f,32C3232)f,32

+ (f,13C1313 + f,31C3113)f,13 + (f,13C1331 + f,31C3131)f,13

= c1(f2,11 + f2

,22 + f2,33

)+ 2c4 (f,11f,22 + f,22f,33 + f,33f,11)

+ 4c7(f2,12 + f2

,23 + f2,31

)

(A.20)

However, since the strain tensor is symmetric, εkl = εlk when k 6= l, it is seen that the number of independentcoefficients is reduced to 3

[C] =

C1111 C1122 C1133 0 0 0C2211 C2222 C2233 0 0 0C3311 C3322 C3333 0 0 0

0 0 0 C1212 0 00 0 0 0 C2323 00 0 0 0 0 C1313

=

c1 c4 c4 0 0 0c4 c1 c4 0 0 0c4 c4 c1 0 0 00 0 0 c7 0 00 0 0 0 c7 00 0 0 0 0 c7

(A.21)

where it is used Voigt notation and symmetric stress tensors

σ4 = σ12 = σ21 = σ7

σ5 = σ23 = σ32 = σ8

σ6 = σ31 = σ13 = σ9

(A.22)

51

The symmetry of the stress and strain tensors allows us to express them as column matrices (or vectors) inthe form

[σ] =

σ11

σ22

σ33

σ12

σ23

σ31

and [ε] =

ε11

ε22

ε33

γ12

γ23

γ31

(A.23)

where γ12 = 2ε12, γ23 = 2ε23 and γ31 = 2ε31 are the engineering shear strains.

[C] =

c1 c4 c4 0 0 0c4 c1 c4 0 0 0c4 c4 c1 0 0 00 0 0 c7 0 00 0 0 0 c7 00 0 0 0 0 c7

(A.24)

which implies that Eq. (A.15) may be written on a more compact form, i.e.

∂f

∂σ

T[C]

∂f

∂σ

=

∂f

∂σiCij

∂f

∂σj

= c1

[(∂f

∂σ1

)2

+

(∂f

∂σ2

)2

+

(∂f

∂σ3

)2]

+ 2c4

[∂f

∂σ1

∂f

∂σ2+

∂f

∂σ2

∂f

∂σ3+

∂f

∂σ3

∂f

∂σ1

]

+ 4c7

[(∂f

∂σ4

)2

+

(∂f

∂σ5

)2

+

(∂f

∂σ6

)2] (A.25)

52

Appendix B. Radial Return Algorithm

As discussed in Section 5.2, the system of equations in Eq.(84) can be extensively simplified and the returnmapping for the von Mises model, with associated flow rule and isotropic hardening, was reduced to a singlenonlinear equation with the incremental plastic multiplier ∆λ as the only unknown. The objective of thissection is to give a theoretical understanding of the result ϕ = ϕtrial − 3G∆λ, which leads to this specialcase called the radial return algorithm. It is readily seen that the radial return algorithm reduce the numberof equations and is important to make the state-update procedure more computationally efficient, and toimprove the performance of the overall finite element scheme. For simplicity only the isotropic strain hard-ening is considered in the constitutive equations.

Starting with the general elastic predictor plastic corrector approach, the stress state at time tn+1 may beexpressed as

σn+1 = σn + C :(∆εn+1 −∆εpn+1

)= σtrialn+1 −C : ∆εpn+1 = σtrialn+1 −∆σp (B.1)

where σtrialn+1 is the elastic predictor moving the stress outside the yield surface, and the plastic corrector∆σp returns the stress to the yield surface (Figure B.31).

Introducing the associated flow rule, Eq.(B.1) reads

σn+1 = σtrialn+1 −3

2∆λn+1C :

σ′

n+1

σeq(B.2)

Using the relation (Appendix C)

C : σ′

n+1 = 2Gσ′

n+1 (B.3)

Eq.(B.2) is given by

σn+1 = σtrialn+1 − 3Gσ′

n+1

σeq∆λn+1 (B.4)

In the case of a three-dimensional stress state, the von Mises yield criterion is a circular cylinder in theprincipal stress space with the deviatoric plane normal to the hydrostatic axis (Figure 4). The von Misesflow vector εp is purely deviatoric, and the hydrostatic stress σH,n+1 is unaffected by the plastic flow(σtrialH,n+1 = σH,n+1). This implies that the stress update can be formulated more efficiently be splitting thestress into deviatoric and hydrostatic parts, where only the deviatoric stress component need to be includedin the return mapping, i.e.

σ′

n+1 = σ′trialn+1 − 3G

σ′

n+1

σeq∆λn+1 = σ

′trialn+1 − 3G

σ′

n+1

ϕ(σ′n+1)

∆λn+1 (B.5)

Rearranging Eq.(B.5) results in an important result, i.e.

σ′

n+1

(1 +

3G∆λn+1

ϕ(σ′n+1)

)= σ

′trialn+1 (B.6)

which implies that the trial and updated deviatoric stresses are co-linear, i.e.

σ′

n+1

ϕ(σ′n+1

) =σ′trialn+1

ϕ(σ′trialn+1

) (B.7)

53

This means that the gradient of the yield surface coincide at the trial and updated stress states (FigureB.31). Substituting the result in Eq.(B.7) into (B.5) results in the following simplification

σ′

n+1 = σ′trialn+1

(1− 3G∆λn+1

ϕ(σ′trialn+1 )

)(B.8)

where ϕ(σ′trialn+1 ) =

√3J2

(σ′trialn+1

)is the elastic trial von Mises equivalent stress.

It is also noted that σ′trialn+1 is a constant vector in the return mapping, and that the deviatoric stress σ

′

n+1

is a function only dependent of ∆λ in Eq.(B.8). Furthermore, Eq.(B.8) implies that, in the radial returnalgorithm with von Mises yielding, the updated deviatoric stress is obtained by scaling the trial deviatoricstress by a factor 1 − 3G∆λn+1/ϕ(σ

′trialn+1 ). The geometric representation of this update formula in the

deviatoric plane is illustrated in Figure B.31.

σ3

σ2σ1

π-plane

∂f∂σ

∣∣∣n+1

σ′

n

σ′trialn+1

σ′

n+1

3G∆λ n

+1

ϕ(σ′ trial

n+

1)σ′ trial

n+

1

surface at tn

surface at tn+1

Figure B.31: The implicit elastic predictor plastic corrector radial return map algorithm for von Misesyielding in the deviatoric plane.

Finally, by inserting Eq.(B.8) into the yield criterion, the following nonlinear scalar equation is obtained

f(∆λ) = ϕ(σ′

n+1

)− σy,n+1

= ϕ(σ′trialn+1

)(1− 3G∆λn+1

ϕ(σ′trialn+1 )

)− σy,n+1

= ϕ(σ′trialn+1

)− 3G∆λn+1 − σy,n+1

= ϕ(σ′trialn+1

)− 3G∆λn+1 − (σy,n +HR,n+1∆λn+1) = 0

(B.9)

54

where the incremental plastic multiplier ∆λ is the only unknown. In Eq.(B.9) it is utilized that ϕ(aσ′) =

aϕ(σ′), since the function ϕ(σ

′) is assumed to be a positive homogeneous function of order one of the stress.

Thus,

ϕ(σ′

n+1) = ϕ

(σ′trialn+1

[1− 3G∆λn+1

ϕ(σ′trialn+1 )

])= ϕ(σ

′trialn+1 )

(1− 3G∆λn+1

ϕ(σ′trialn+1 )

) (B.10)

By using Eq.(B.9), the incremental plastic multiplier at time tn+1 is expressed as

∆λn+1 =ϕ(σ′trialn+1

)− σy,n

3G∆λn+1 +HR,n+1(B.11)

which, in the case of nonlinear hardening, could be solved by a Newton-Raphson method

∆λi+1n+1 = ∆λin+1 −

f(∆λ)

f ′(∆λ)

= ∆λin+1 −ϕ(σ′trialn+1

)− 3G∆λin+1 − σy,n+1

−3G−HR

(B.12)

where i is the local iteration counter. For linear hardening ∆λ is found directly and no local iterations arenecessary.

Finally, the stresses and strains are updated according to Eq.(B.2)

σn+1 = σtrialn+1 −3

2∆λn+1C :

σ′

n+1

σeq

= σtrialn+1 −3

2∆λn+1C :

σ′trialn+1

ϕ(σ′trialn+1

)pn+1 = pn + ∆λn+1

εpn+1 = εpn +3

2

σ′trialn+1

ϕ(σ′trialn+1

)∆λn+1

(B.13)

55

Appendix C. Isotropic elastic material

Assuming isotropic linear elastic behavior of the material reduces the stiffness tensor to only depend on theLame constants, i.e

Cijkl = λelδijδkl + µel (δikδjl + δilδjk) (C.1)

and the stress is then given by the linear relation

σij = Cijklεkl = λelδijδklεkl + µel (δikδjlεkl + δilδjkεkl)

= λelδijεkk + µel (εij + εji)

= λelδijεkk + 2µel (εij)

(C.2)

where the following relations are used

δikδjlεkl = δikεjk = εij

δilδjkεkl = δilεlj = εji

εij = εji

(C.3)

By inserting the expression for the Lame constants, Eq.(C.1) reads

Cijkl =νE

(1 + ν)(1− 2ν)δijδkl +

E

2(1 + ν)(δikδjl + δilδjk)

=E

1 + ν

[1

2(δikδjl + δilδjk) +

ν

1− 2νδijδkl

]= 2G

[1

2(δikδjl + δilδjk) +

ν

1− 2νδijδkl

] (C.4)

and by multiplying the isotropic elasticity tensor in Eq.(C.4) with the deviatoric part of the stress tensor,this product reduces to

Cijklσ′

kl = 2G

[1

2

(δikδjlσ

′

kl + δilδjkσ′

kl

)+

ν

1− 2νδijδklσ

′

kl

]= 2G

[1

2

(σ′

ij + σ′

ji

)+

ν

1− 2νδijσ

′

kk

] (C.5)

Utilizing the symmetry of the stress tensor, i.e. σ′

ij = σ′

ji, and the definition of the deviatoric stress

σ′

ij = σij −1

3σkkδij , σ

′

ii = σii −1

3σkkδii = 0 (C.6)

Eq.(C.5) reduces to the following expression

Cijklσ′

kl = 2Gσ′

ij (C.7)

which is a useful relation in the radial return map algorithm.

56

Appendix D. Two-dimensional stress analysis

This section is an extened version of Section 3 and shows useful relationships when dealing with the particularcase of plane stress and three shear stress components. Plane stress is already defined in Section 3 as thestate of stress in which the normal stress σ33 and the shear stresses σ13,23 are assumed to be zero [11]. Froma practical point of view this assumption is typically introduced in the analysis of bodies in which one ofthe dimensions (i.e. the thickness) is much smaller than the others and is subjected to loads that result indominant stresses perpendicular to the thickness direction [5]. However, the same assumption may also beused when only the normal stress σ33 is assumed to be zero. Thus, all the shear components are consideredin the calculation.

Appendix D.1. Two-dimensional theory of elasticity

Recall that, in the particular case of plane stress, the linear elastic stress-strain compliance relationship foran isotropic material can be written as a system of equations in matrix form given as

ε11

ε22

ε33

2ε12

2ε23

2ε31

= Sσ =1

E

1 −ν −ν 0 0 0−ν 1 −ν 0 0 0−ν −ν 1 0 0 00 0 0 2(1 + ν) 0 00 0 0 0 2(1 + ν) 00 0 0 0 0 2(1 + ν)

σ11

σ22

0σ12

σ23

σ31

(D.1)

The normal stress σ33 which is assumed to be zero in the stress vector indicates that its associated columnin the compliance matrix S can be neglected (i.e. column 3). Furthermore, if we neglect the row associatedwith the strain components in the thickness direction εi3 (i = 1, 2, 3), the stress-strain compliance relation-ship reduces to a 5× 5 matrix

ε11

ε22

2ε12

2ε23

2ε31

= Sσ =1

E

1 −ν 0 0 0−ν 1 0 0 00 0 2(1 + ν) 0 00 0 0 2(1 + ν) 00 0 0 0 2(1 + ν)

σ11

σ22

σ12

σ23

σ31

(D.2)

Then, the elasticity matrix C for plane stress is found by inverting the plane stress compliance matrix Sand given by Hooke’s law as

σ11

σ22

σ12

σ23

σ31

= Cε =E

1− ν2

1 ν 0 0 0ν 1 0 0 00 0 1−ν

2 0 00 0 0 1−ν

2 00 0 0 0 1−ν

2

ε11

ε22

2ε12

2ε23

2ε31

(D.3)

It is important to note that the elasticity matrix for plane stress is not found by removing columns and rowsfrom the general isotropic elasticity matrix in Eq. (C.1).

It should also be noted that the procedure in finding the principal stresses σi of the stress tensor σ are nowdetermined by using Eq. (57), i.e.

σi = σ′

i + σH = σ′

i +1

3tr(σ) (D.4)

where the major principal stress σ1 is used in the Cockcroft-Latham failure criterion in Eq. (46). Thus, theprocedure in Section 3.1 is not applicable when using all the shear components.

57

Appendix E. Implementation of Return Map Algorithm in EUROPLEXUS

This appendix shows a somewhat more detailed version of the numerical scheme given in Section 5.1. Withinthe time increment ∆tn+1 the scheme may then be explained as follows:

1. Set the initial values of internal variables to the converged values of the previous step at tn, andcalculate the speed of sound. Note that the number of shear stress components (ITAU ) and planestress conditions (IPLANC ) (provided by the element routine) are used to SELECT the appropriatestress state in the material routine.

Governing equations

Define useful parameters related to the theory ofelasticity

T1 = 1 + ν, T2 = 1− 2ν, T3 = 1− ν

Compute the speed of sound based on thestress state, i.e.,IF 3D stress state THEN

CSON =

√E(1− ν)

ρ(1 + ν)(1− 2ν)

ELSEIF plane stress THEN

CSON =

√E

ρ(1− ν2)

ELSEIF uniaxial stress THEN

CSON =

√E

ρ

Implementation in EUROPLEXUS

SUBROUTINE VPJC (ITAU, IPLANC, PROPS, STRAININC, STRESSOLD,

> STRAIN, STATE, STRESSNEW, CSON, FAIL_FLAG)

* sound speed

T1 = R1 + POISS

T2 = R1 - R2*POISS

T3 = R1 - POISS

*

SELECT CASE (IPLANC)

CASE (0) ! NO PLANE STRESS CONDITION

CSON = SQRT (YOUNG*T3 / (RHO*T1*T2))

CASE (1) ! PLANE STRESS CONDITION (SIG(3)=0)

CSON = SQRT (YOUNG / (RHO*T1*T3))

CASE (2) ! UNIAXIAL STRESS CONDITION (SIG(2)=SIG(3)=0)

CSON = SQRT (YOUNG / RHO)

CASE DEFAULT ! CHECK THIS ONLY HERE

CALL ERRMSS (’VPJC’, ’WRONG IPLANC’)

STOP ’VPJC - WRONG IPLANC’

END SELECT

*

2. Compute the elasticity tensor.

Goverening equations

IF a 3D stress state is used THEN use Eqs. (3)-(4)

C = λelI ⊗ I + 2µelI =

C1 C4 C5 0 0 0C5 C2 C4 0 0 0C5 C6 C3 0 0 00 0 0 C7 0 00 0 0 0 C8 00 0 0 0 0 C9

C1 = C2 = C3 =

(1 − ν)E

(1 + ν)(1 − 2ν), C4 = C5 = C6 =

νC1

1 − ν,

C7 = C8 = C9 =1

2

E

1 + ν

ELSEIF a 2D stress state is used THEN useEq. (60)

C =

C1 C4 0C5 C2 00 0 C7

C1 = C2 =

E

1 − ν2, C4 = C5 = νC1, C7 =

1

2

E

1 + ν

ELSE a 1D stress state is used THEN use

C =

C1 0 00 0 00 0 0

, C1 = E

Implementation in EUROPLEXUS

* compute the elasticity tensor

SELECT CASE (IPLANC)

CASE (0) ! NO PLANE STRESS CONDITION

C1 = YOUNG*T3 / (T1*T2)

C4 = POISS*C1 / T3

CASE (1) ! PLANE STRESS CONDITION (SIG(3)=0)

C1 = YOUNG / (T1*T3)

C4 = POISS*C1

CASE (2) ! UNIAXIAL STRESS CONDITION (SIG(2)=SIG(3)=0)

C1 = YOUNG

C4 = R0 ! THESE VALUES HAVE TO BE CHECKED

END SELECT

*

C2 = C1

C3 = C1

C5 = C4

C6 = C4

C7 = R5*YOUNG / T1

C8 = C7

C9 = C7

*

58

3. & 4. Use the total stain increment ∆εn+1 from the global equilibrium to compute the elastic predictor.Then, compute the von Mises equivalent stress in terms of the elastic trial stress and Eq. (35). Notethat Eqs.(35) and (34) are used to calculate the gradients, i.e., σ12 = σ21, σ23 = σ32 and σ13 = σ31.

Goverening equations

σtrial11,n+1 = σ11,n + C1∆ε11,n+1 + C4∆ε22,n+1

+ C5∆ε33,n+1

IF a 3D stress state is used THEN

σtrial22,n+1 = σ22,n + C4∆ε11,n+1 + C2∆ε22,n+1

+ C6∆ε33,n+1

σtrial33,n+1 = σ33,n + C5∆ε11,n+1 + C6∆ε22,n+1

+ C3∆ε33,n+1

σtrial12,n+1 = σ12,n + C7∆ε12,n+1

σtrial23,n+1 = σ23,n + C8∆ε23,n+1

σtrial31,n+1 = σ31,n + C9∆ε31,n+1

3

2σ′trialn+1 : σ

′trialn+1 = (σ

trial11,n+1)

2+ (σ

trial22,n+1)

2

+ (σtrial33,n+1)

2 − σtrial11,n+1σtrial22,n+1

+ σtrial22,n+1σ

trial33,n+1 − σ

trial33,n+1σ

trial11,n+1

+ 3((σtrial12,n+1)

2+ (σ

trial23,n+1)

2

+ (σtrial31,n+1)

2)ELSEIF a 2D plane stress state is used THEN

σtrial22,n+1 = σ22,n + C4∆ε11,n+1 + C2∆ε22,n+1

+ C6∆ε33,n+1

σtrial33,n+1 = 0

σtrial12,n+1 = σ12,n + C7∆ε12,n+1

3

2σ′trialn+1 : σ

′trialn+1 = (σ

trial11,n+1)

2+ (σ

trial22,n+1)

2

− σtrial11,n+1σtrial22,n+1 + 3

((σtrial12,n+1)

2)ELSE a 1D stress state is used and

σtrial22,n+1 = 0

σtrial33,n+1 = 0

3

2σ′trialn+1 : σ

′trialn+1 = (σ

trial11,n+1)

2

END IF

σtrialeq,n+1 = ϕ(σ

trialn+1 ) =

√3

2σ′trialn+1

: σ′trialn+1

Implementation in EUROPLEXUS

* compute the elastic trial stress and the trial equivalent stress

*

STRESS(1) = STRESSOLD(1) + C1*STRAININC(1) + C4*STRAININC(2)

> + C5*STRAININC(3)

*

SELECT CASE (IPLANC) ! TREAT NORMAL COMPONENTS FIRST

CASE (0) ! NO PLANE STRESS CONDITION

STRESS(2) = STRESSOLD(2) + C4*STRAININC(1) + C2*STRAININC(2)

> + C6*STRAININC(3)

STRESS(3) = STRESSOLD(3) + C5*STRAININC(1) + C6*STRAININC(2)

> + C3*STRAININC(3)

PHITRIAL = STRESS(1)*STRESS(1) + STRESS(2)*STRESS(2)

> + STRESS(3)*STRESS(3) - STRESS(1)*STRESS(2)

> - STRESS(2)*STRESS(3) - STRESS(3)*STRESS(1)

CASE (1) ! PLANE STRESS CONDITION (STRESS(3)=0)

STRESS(2) = STRESSOLD(2) + C4*STRAININC(1) + C2*STRAININC(2)

> + C6*STRAININC(3)

STRESS(3) = R0

PHITRIAL = STRESS(1)*STRESS(1) + STRESS(2)*STRESS(2)

> - STRESS(1)*STRESS(2)

CASE (2) ! UNIAXIAL STRESS CONDITION (STRESS(2)=STRESS(3)=0)

STRESS(2) = R0

STRESS(3) = R0

PHITRIAL = STRESS(1)*STRESS(1)

END SELECT

*

SELECT CASE (ITAU) ! ADD SHEAR COMPONENTS IF PRESENT

CASE (0) ! NO SHEAR STRESS IS PRESENT

CASE (1) ! ONLY TAU_XY (STRESS(4)) IS PRESENT

STRESS(4) = STRESSOLD(4) + C7*STRAININC(4) ! THE GAMMAS ARE USED!

PHITRIAL = PHITRIAL + R3*(STRESS(4)*STRESS(4))

CASE (3) ! ALL THREE SHEAR STRESSES (STRESS(4:6)) ARE PRESENT

STRESS(4) = STRESSOLD(4) + C7*STRAININC(4) ! SHEAR STRAINS

STRESS(5) = STRESSOLD(5) + C8*STRAININC(5) ! ARE THE GAMMAS, NOT

STRESS(6) = STRESSOLD(6) + C9*STRAININC(6) ! THE EPSILONS

PHITRIAL = PHITRIAL + R3*(STRESS(4)*STRESS(4)

> + STRESS(5)*STRESS(5)

> + STRESS(6)*STRESS(6))

CASE DEFAULT ! WE CHECK THIS ONLY IN THE FIRST SELECT CASE ON ITAU

CALL ERRMSS (’VPJC’, ’WRONG ITAU’)

STOP ’VPJC - WRONG ITAU’

END SELECT

*

PHITRIAL = SQRT (PHITRIAL)

*

59

5. Define initial values of internal variables, to be used in return mapping algorithm, to the convergedvalues of the previous step at tn. Check if temperature softening is included by the user (i.e., m 6= 0in material input), compute a value for the yield function f(σtrialn+1 ) and set the incremental plasticmultiplier ∆λn+1 equal to zero.

Goverening equations

pn = STATE(3)

Rn = STATE(6)

σy,n = A + Rn

Wn = STATE(14)

IF temperature softening is specified in materialinput (m 6= 0) THEN

T∗

=Tn − Tr

Tm − Tr

Γp(Tn) =(1 − T∗m

)fn+1 = σ

trialeq,n+1 − σy,nΓp(Tn)

ELSE m = 0 and the CPU time can be reducedby skipping the softening term

fn+1 = σtrialeq,n+1 − σy,n

ENDIF

∆λn+1 = 0

δλn+1 = 0

IF ϕtrialn+1 = 0 THEN

ϕn+1 = 1

ELSE

ϕn+1 = ϕtrialn+1 = σ

trialeq,n+1

ENDIF

Implementation in EUROPLEXUS

*--------------------------------------------------------------

PN = STATE(3)

RP = STATE(6)

SIGMAY = SY + RP

*

* import old value damage variables

WE = STATE(14)

*

IF (TEMP_SOFT) THEN

TSTAR = (T-TR) / (TM-TR)

TSOFT = R1 - TSTAR**M

F = PHITRIAL - SIGMAY*TSOFT

ELSE ! M = 0 --> THE CODE ASSUMES TSTAR=0, SO THAT TSOFT=1

F = PHITRIAL - SIGMAY

ENDIF

DLAMBDA = R0

DDLAMBDA = R0

IF (PHITRIAL == R0) THEN

PHI = R1

ELSE

PHI = PHITRIAL

ENDIF

*

60

6. Check for plastic admissibility.

Goverening equations

IF f(σtrialn+1 ) > 0 THEN apply return mapping by using the

cutting plane algorithm (SOLU = 1) to find the incremental

change in the plastic multiplier ∆λn+1.

Superscript i denotes the local iteration counter.

Implementation in EUROPLEXUS

IF (F > R0) THEN

*

* plastic step: apply return mapping

* the cutting plane method uses values from the previous

* iteration to update the plastic parameter, dlambda.

* thus, all subsequent values (until dlambda) are related to

* the previous iteration, "i".

*--------------------------------------------------------------

P = PN

LOOP_NR : DO NRITER = 1, MXITER

(i) Compute the normal nn+1 to the yield surface based on the initial (i.e., trial) or previous stressconfiguration. Note that Eqs.(35) and (34) are used to calculate the gradients, i.e., σ12 = σ21,σ23 = σ32 and σ13 = σ31.

Goverening equations

IF a 3D stress state is used THEN

n11 = ∂f/∂σ11 = (σ11 − 0.5(σ22 + σ33))/σeq

n22 = ∂f/∂σ22 = (σ22 − 0.5(σ33 + σ11))/σeq

n33 = ∂f/∂σ33 = (σ33 − 0.5(σ11 + σ33))/σeq

n12 = ∂f/∂σ12 = 3σ12/σeq

n23 = ∂f/∂σ23 = 3σ23/σeq

n31 = ∂f/∂σ31 = 3σ31/σeq

ELSEIF a plane stress state is used THEN

n11 = ∂f/∂σ11 = (σ11 − 0.5(σ22))/σeq

n22 = ∂f/∂σ22 = (σ22 − 0.5(σ11))/σeq

n33 = 0

n12 = ∂f/∂σ12 = 3σ12/σeq

ELSE a 1D stress state is used and

n11 = ∂f/∂σ11 = σ11/σeq

Implementation in EUROPLEXUS

* compute the respective gradients of the von mises yield surface

RNV(2:6) = R0 ! INITIALIZE TO ZERO "OPTIONAL" COMPONENTS

SELECT CASE (IPLANC) ! TREAT NORMAL COMPONENTS FIRST

CASE (0) ! NO PLANE STRESS CONDITION

RNV(1) = (STRESS(1) - R5*(STRESS(2) + STRESS(3))) / PHI

RNV(2) = (STRESS(2) - R5*(STRESS(3) + STRESS(1))) / PHI

RNV(3) = (STRESS(3) - R5*(STRESS(1) + STRESS(2))) / PHI

CASE (1) ! PLANE STRESS CONDITION (STRESS(3)=0)

RNV(1) = (STRESS(1) - R5*STRESS(2)) / PHI

RNV(2) = (STRESS(2) - R5*STRESS(1)) / PHI

CASE (2) ! UNIAXIAL STRESS CONDITION (STRESS(2)=STRESS(3)=0)

RNV(1) = STRESS(1) / PHI ! TO BE CHECKED ...

END SELECT

*

SELECT CASE (ITAU) ! TREAT SHEAR COMPONENTS

CASE (0) ! NO SHEAR STRESS IS PRESENT

CASE (1) ! ONLY TAU_XY (STRESS(4)) IS PRESENT

RNV(4) = R3*STRESS(4) / PHI

CASE (3) ! ALL THREE SHEAR STRESSES (STRESS(4:6)) ARE PRESENT

RNV(4) = R3*STRESS(4) / PHI

RNV(5) = R3*STRESS(5) / PHI

RNV(6) = R3*STRESS(6) / PHI

END SELECT

(ii) Compute hardening modulus HR,n+1 according to Eq. (42)

Goverening equations

HiR,n+1 = Q1C1 exp(−C1p

in+1)+Q2C2 exp(−C2p

in+1)

Implementation in EUROPLEXUS

* compute hardening modulus

HR = QR1*CR1*EXP(-CR1*P) + QR2*CR2*EXP(-CR2*P)

61

(iii) Compute denominator in Eq. (90) using Eqs. (91), (92), (93) and (98)

Goverening equations

IF a 3D stress state is used THEN

nTCn = C1

[(n11)

2+ (n22)

2+ (n33)

2]

+ 2C4[n11n22 + n22n33 + n33n11]

+ 4C7

[(n12)

2+ (n23)

2+ (n31)

2]

ELSEIF a plane stress state is used THEN

nTCn = C1

[(n11)

2+ (n22)

2]

+ 2C4[n11n22] + 4C7

[(n12)

2]

ELSE a 1D stress state is used and

nTCn = C1 (n11)

2

υip,n+1 =

1 +∆λin+1

p0∆tn+1

C

∂υip,n+1

∂∆λin+1

=C

p0∆tn+1

1 +∆λin+1

p0∆tn+1

C−1

IF temperature softening is activated (m 6= 0)THEN

Γp(Tn) =

(1 −

[Tn − Tr

Tm − Tr

]m)

DENOM = nTCn + υ

ip,n+1Γp,nH

iR,n+1

+ σiy,n+1Γp,n

∂υip,n+1

∂∆λin+1

ELSE m = 0 and skip the softening term

DENOM = nTCn + υ

ip,n+1H

iR,n+1

+ σiy,n+1

∂υip,n+1

∂∆λin+1

ENDIF

Implementation in EUROPLEXUS

* compute denominator in the residual derivative of dlambda

SELECT CASE (IPLANC) ! TREAT NORMAL COMPONENTS FIRST

CASE (0) ! NO PLANE STRESS CONDITION

DENOMCP = C1*(RNV(1)*RNV(1) + RNV(2)*RNV(2) + RNV(3)*RNV(3))

> + R2*C4*(RNV(1)*RNV(2) + RNV(2)*RNV(3) + RNV(3)*RNV(1))

CASE (1) ! PLANE STRESS CONDITION (RNV(3)=0)

DENOMCP = C1*(RNV(1)*RNV(1) + RNV(2)*RNV(2))

> + R2*C4*(RNV(1)*RNV(2))

CASE (2) ! UNIAXIAL STRESS CONDITION (RNV(2)=RNV(3)=0)

DENOMCP = C1*(RNV(1)*RNV(1)) ! TO BE CHECKED ...

END SELECT

*

SELECT CASE (ITAU) ! TREAT SHEAR COMPONENTS

CASE (0) ! NO SHEAR STRESS IS PRESENT

CASE (1) ! ONLY TAU_XY (RNV(4)) IS PRESENT

DENOMCP = DENOMCP + R4*C7*(RNV(4)*RNV(4))

CASE (3) ! ALL THREE SHEAR STRESSES (RNV(4:6)) ARE PRESENT

DENOMCP = DENOMCP + R4*C7*(RNV(4)*RNV(4) + RNV(5)*RNV(5)

> + RNV(6)*RNV(6))

END SELECT

*

DENOMV = (R1 + DLAMBDA / (PDOT*DT))**C

DENOMDV = (C / (PDOT*DT))*(R1 + DLAMBDA / (PDOT*DT))**(C - R1)

IF (TEMP_SOFT) THEN

TSTAR = (T - TR) / (TM - TR)

TSOFT = R1 - TSTAR**M

DENOM = DENOMCP + HR*DENOMV*TSOFT + SIGMAY*DENOMDV*TSOFT

ELSE ! M = 0 --> THE CODE ASSUMES TSTAR=0, SO THAT TSOFT=1

DENOM = DENOMCP + HR*DENOMV + SIGMAY*DENOMDV

ENDIF

(iv) Compute incremental change in the plastic multiplier δλi+1n+1 in Eq. (90) and update variable

∆λi+1n+1

Goverening equations

δλi+1n+1

=fin+1

DENOM

∆λi+1n+1

= ∆λin+1 + δλ

i+1n+1

Implementation in EUROPLEXUS

* compute residual increment ddlambda and update variable dlambda

DDLAMBDA = F / DENOM

DLAMBDA = DLAMBDA + DDLAMBDA

62

(v) Update internal variables dependent on ∆λi+1n+1

Goverening equations

pi+1n+1

= pin+1 + δλ

i+1n+1

Ri+1n+1

= Rn +HiR,n+1∆λ

i+1n+1

σi+1y,n+1

= A + Ri+1n+1

Implementation in EUROPLEXUS

* update internal variables dependent on dlambda

P = P + DDLAMBDA

R = RP + HR*DLAMBDA

SIGMAY = SY + R

(vi) Update stress components using the elastic Hooke’s law according to Eq. (83). Note that thematerial routine uses a a fully vectorized representation of the stress and strain vector, i.e., σ =[σ11, σ22, σ33, σ12, σ23, σ31]T and ε = [ε11, ε22, ε33, γ12, γ23, γ31]T = [ε11, ε22, ε33, 2ε12, 2ε23, 2ε31]T

Goverening equations

IF a 3D stress state is used THEN

σi+111,n+1

= σi11,n+1 − δλ

i+1n+1

(C1n11 + C4n22 + C5n33)

σi+122,n+1

= σi22,n+1 − δλ

i+1n+1

(C4n11 + C2n22 + C6n33)

σi+133,n+1

= σi33,n+1 − δλ

i+1n+1

(C5n11 + C6n22 + C3n33)

σi+112,n+1

= σi12,n+1 − δλ

i+1n+1

C7n12

σi+123,n+1

= σi23,n+1 − δλ

i+1n+1

C8n23

σi+131,n+1

= σi31,n+1 − δλ

i+1n+1

C9n31

ELSEIF a plane stress state is used THEN

σi+111,n+1

= σi11,n+1 − δλ

i+1n+1

(C1n11 + C4n22 + C5n33)

σi+122,n+1

= σi22,n+1 − δλ

i+1n+1

(C4n11 + C2n22 + C6n33)

σi+133,n+1

= 0

σi+112,n+1

= σi12,n+1 − δλ

i+1n+1

C7n12

ELSE a 1D stress state is used and

σi+111,n+1

= σi11,n+1 − δλ

i+1n+1

C1n11

Implementation in EUROPLEXUS

* update stress components using the elastic (hooke’s) law

SELECT CASE (IPLANC) ! TREAT NORMAL COMPONENTS FIRST

CASE (0) ! NO PLANE STRESS CONDITION

STRESS(1) = STRESS(1) -

> DDLAMBDA*(C1*RNV(1) + C4*RNV(2) + C5*RNV(3))

STRESS(2) = STRESS(2) -

> DDLAMBDA*(C4*RNV(1) + C2*RNV(2) + C6*RNV(3))

STRESS(3) = STRESS(3) -

> DDLAMBDA*(C5*RNV(1) + C6*RNV(2) + C3*RNV(3))

CASE (1) ! PLANE STRESS CONDITION (RNV(3)=0)

STRESS(1) = STRESS(1) -

> DDLAMBDA*(C1*RNV(1) + C4*RNV(2))

STRESS(2) = STRESS(2) -

> DDLAMBDA*(C4*RNV(1) + C2*RNV(2))

CASE (2) ! UNIAXIAL STRESS CONDITION (RNV(2)=RNV(3)=0)

STRESS(1) = STRESS(1) - DDLAMBDA*C1*RNV(1) ! TO BE CHECKED ...

END SELECT

*

SELECT CASE (ITAU) ! TREAT SHEAR COMPONENTS

CASE (0) ! NO SHEAR STRESS IS PRESENT

CASE (1) ! ONLY TAU_XY (RNV(4)) IS PRESENT

STRESS(4) = STRESS(4) - DDLAMBDA*C7*RNV(4)

CASE (3) ! ALL THREE SHEAR STRESSES (RNV(4:6)) ARE PRESENT

STRESS(4) = STRESS(4) - DDLAMBDA*C7*RNV(4)

STRESS(5) = STRESS(5) - DDLAMBDA*C8*RNV(5)

STRESS(6) = STRESS(6) - DDLAMBDA*C9*RNV(6)

END SELECT

*

63

(vii) Compute the updated von Mises equivalent stress using Eq. (35)

Goverening equations

IF a 3D stress state is used THEN

3

2σ′,i+1n+1

: σ′,i+1n+1

= (σi+111,n+1

)2

+ (σi+122,n+1

)2

+ (σi+133,n+1

)2 − σi+1

11,n+1σi+122,n+1

+ σi+122,n+1

σi+133,n+1

− σi+133,n+1

σi+111,n+1

+ 3((σi+112,n+1

)2

+ (σi+123,n+1

)2

+ (σi+131,n+1

)2)

ELSEIF a plane stress state is used THEN

3

2σ′,i+1n+1

: σ′,i+1n+1

= (σi+111,n+1

)2

+ (σi+122,n+1

)2

− σi+111,n+1

σi+122,n+1

+ 3((σi+112,n+1

)2)

ELSE a 1D stress state is used and

3

2σ′,i+1n+1

: σ′,i+1n+1

= (σi+111,n+1

)2

ENDIF

σi+1eq,n+1

= ϕ(σi+1n+1

) =

√3

2σ′,i+1n+1

: σ′,i+1n+1

Implementation in EUROPLEXUS

* compute updated equivalent stress

SELECT CASE (IPLANC) ! TREAT NORMAL COMPONENTS FIRST

CASE (0) ! NO PLANE STRESS CONDITION

PHI = STRESS(1)*STRESS(1) + STRESS(2)*STRESS(2)

> + STRESS(3)*STRESS(3) - STRESS(1)*STRESS(2)

> - STRESS(2)*STRESS(3) - STRESS(3)*STRESS(1)

CASE (1) ! PLANE STRESS CONDITION (STRESS(3)=0)

PHI = STRESS(1)*STRESS(1) + STRESS(2)*STRESS(2)

> - STRESS(1)*STRESS(2)

CASE (2) ! UNIAXIAL STRESS CONDITION (STRESS(2)=STRESS(3)=0)

PHI = STRESS(1)*STRESS(1)

END SELECT

*

SELECT CASE (ITAU) ! ADD SHEAR COMPONENTS IF PRESENT

CASE (0) ! NO SHEAR STRESS IS PRESENT

CASE (1) ! ONLY TAU_XY (STRESS(4)) IS PRESENT

PHI = PHI + R3*(STRESS(4)*STRESS(4))

CASE (3) ! ALL THREE SHEAR STRESSES (STRESS(4:6)) ARE PRESENT

PHI = PHI + R3*(STRESS(4)*STRESS(4)

> + STRESS(5)*STRESS(5)

> + STRESS(6)*STRESS(6))

END SELECT

*

PHI = SQRT (PHI)

(viii) Update the yield function in Eq. (84) according to the new stress state computed in (vi) and(vii). Check for convergence.

Goverening equations

IF temperature softening (m 6= 0) THEN

fi+1v,n+1

= ϕi+1n+1

− σi+1y,n+1

(•)

(•)

=

1 +∆λ

i+1n+1

p0∆tn+1

C (

1 −[Tn − Tr

Tm − Tr

]m)

RESNOR =

∣∣∣∣∣∣∣fv(∆λ

i+1n+1

)

σi+1y,n+1

υi+1p,n+1

Γp

∣∣∣∣∣∣∣ELSE m = 0 and skip the softening term

fi+1v,n+1

= ϕi+1n+1

− σi+1y,n+1

(•)

(•)

=

1 +∆λ

i+1n+1

p0∆tn+1

C

RESNOR =

∣∣∣∣∣∣∣fv(∆λ

i+1n+1

)

σi+1y,n+1

υi+1p,n+1

∣∣∣∣∣∣∣ENDIF

IF RESNOR < TOL THEN continue to 7.

ELSE i = i+ 1 and GO TO (i)

Implementation in EUROPLEXUS

* compute yield function and new residual

IF (TEMP_SOFT) THEN

TSTAR = (T - TR) / (TM - TR) ! THIS RECALCULATION OF TSTAR

TSOFT = R1 - TSTAR**M ! AND TSOFT SEEMS REDUNDANT HERE

F = PHI - SIGMAY*((R1 + DLAMBDA / (PDOT*DT))**C)*TSOFT

*

* check convergence

RESNOR = F / (SIGMAY*((R1 + DLAMBDA / (PDOT*DT))**C)*TSOFT)

RESNOR = ABS (RESNOR)

ELSE ! M = 0 --> THE CODE ASSUMES TSTAR=0, SO THAT TSOFT=1

F = PHI - SIGMAY*((R1 + DLAMBDA / (PDOT*DT))**C)

* check convergence

RESNOR = F / (SIGMAY*((R1 + DLAMBDA / (PDOT*DT))**C))

RESNOR = ABS (RESNOR)

ENDIF

*

IF (RESNOR <= TOL) THEN

64

7. Update the temperature according to the low coupling approach using Eq. (45) and erode element ifT > Tm

Goverening equations

Tn+1 = Tn +χσeq,n+1∆λn+1

ρCT

Implementation in EUROPLEXUS

* update temperature (only if temperature softening is activated)

IF (TEMP_SOFT) THEN

T = T + TQ*PHI*DLAMBDA / (RHO*CP)

IF (T >= TM) THEN ! CHECK IF MELTING OCCURS

STATE(13) = TM

D = DC ! this will cause the GP to fail (see 990)

GO TO 990

ENDIF

ENDIF

8. Compute the hydrostatic stress according to Eq. (32)

Goverening equations

σH,n+1 =1

3tr(σn+1) =

1

3(σ11,n+1+σ22,n+1+σ33,n+1)

Implementation in EUROPLEXUS

* compute hydrostatic stress based on new stress state,

SIGMAH = STRESS(1)

SELECT CASE (IPLANC)

CASE (0) ! NO PLANE STRESS CONDITION

SIGMAH = SIGMAH + STRESS(2) + STRESS(3)

CASE (1) ! PLANE STRESS CONDITION (STRESS(3)=0)

SIGMAH = SIGMAH + STRESS(2)

CASE (2) ! UNIAXIAL STRESS CONDITION (STRESS(2)=STRESS(3)=0)

END SELECT

*

SIGMAH = SIGMAH / R3

9. Computing the principal stresses σi,n+1 based on the principal deviatoric stresses σ′

i,n+1 as shown inEqs. (55) and (57)

Goverening equations

σ′11,n+1 = σ11,n+1 − σH,n+1

σ′22,n+1 = σ22,n+1 − σH,n+1

σ′33,n+1 = σ33,n+1 − σH,n+1

J2,n+1 =1

2

((σ′11,n+1)

2+ (σ

′22,n+1)

2+ (σ

′33,n+1)

2)

+ (σ12,n+1)2

+ (σ23,n+1)2

+ (σ31,n+1)2

J3,n+1 = σ′11,n+1(σ23,n+1)

2+ σ′22,n+1(σ31,n+1)

2

+σ′33,n+1(σ12,n+1)

2 − σ′11,n+1σ

′22,n+1 + σ

′33,n+1

− 2σ12,n+1σ23,n+1σ31,n+1

θ = arccos

J3,n+1

2

√J32,n+1

/27

, 0 ≤ θ ≤ π

σ′1,n+1 = 2

√J2,n+1

3cos

θ

3

σ′2,n+1 = 2

√J2,n+1

3cos

(θ

3−

2π

3

)

σ′3,n+1 = 2

√J2,n+1

3cos

(θ

3+

2π

3

)

σ1,n+1 = σH,n+1 + σ′1,n+1

σ2,n+1 = σH,n+1 + σ′2,n+1

σ3,n+1 = σH,n+1 + σ′3,n+1

Implementation in EUROPLEXUS

* 3d stress state and compute deviatoric stress

S1 = STRESS(1) - SIGMAH

S2 = STRESS(2) - SIGMAH

S3 = STRESS(3) - SIGMAH

*

* finding principal stress based on the deviatoric stress and invariants

A1 = R5*(S1*S1 + S2*S2 + S3*S3)

*

SELECT CASE (ITAU) ! ADD SHEAR COMPONENTS IF PRESENT

CASE (0) ! NO SHEAR STRESS IS PRESENT

A2 = -S1*S2*S3 ! DETERMINANT OF THE DEVIATORIC STRESS TENSOR

CASE (1) ! ONLY TAU_XY (STRESS(4)) IS PRESENT

A1 = A1 + STRESS(4)*STRESS(4)

A2 = S3*STRESS(4)*STRESS(4) - S1*S2*S3

CASE (3) ! ALL THREE SHEAR STRESSES (STRESS(4:6)) ARE PRESENT

A1 = A1 +

> STRESS(4)*STRESS(4) + STRESS(5)*STRESS(5) +

> STRESS(6)*STRESS(6)

A2 = S1*STRESS(5)*STRESS(5) +

> S2*STRESS(6)*STRESS(6) +

> S3*STRESS(4)*STRESS(4) -

> S1*S2*S3 - R2*STRESS(4)*STRESS(5)*STRESS(6)

END SELECT

*

A3 = -SQRT (R27 / A1)*A2*R5 / A1

A3 = SIGN (MIN (ABS (A3), R1), A3)

A4 = ACOS (A3) / R3

*

* compute (chronological) principal stresses

AUX = SQRT (A1 / R3)

S1 = SIGMAH + R2*AUX*COS(A4)

S2 = SIGMAH + R2*AUX*COS(A4-R2*PI/R3) ! only for postprocessing

S3 = SIGMAH + R2*AUX*COS(A4+R2*PI/R3) ! only for postprocessing

*

65

10. Compute the damage parameter Dn+1 by using Eq. (46), i.e. integrating the major principal stressduring the entire equivalent plastic strain path

Goverening equations

Wn+1 = Wn + max(σ1,n+1, 0)∆λn+1

Dn+1 =Wn+1

Wc

Implementation in EUROPLEXUS

*

WE = WE + MAX (S1, R0)*DLAMBDA

D = WE / WC

*

11. Update state variables to be returned to the element routine in EUROPLEXUS and check for elementerosion.

Goverening equations

STATE(1) = σH,n+1

STATE(2) = σeq,n+1

STATE(3) = pn+1

STATE(4) = σtrialeq,n+1

STATE(5) = fn+1

STATE(6) = Rn+1

STATE(7) = δλn+1

STATE(8) = ∆λn+1

STATE(9) = NRITER

STATE(10) = pn+1 = ∆λn+1/∆tn+1

STATE(11) = Dn+1

STATE(13) = Tn+1

STATE(14) = Wn+1

STATE(15) = CSONn+1

STATE(16) = σ1,n+1

STATE(17) = σ2,n+1

STATE(18) = σ3,n+1

IF Dn+1 > DC THEN

failure and the material point stiffness is deleted.

STATE(12) = 1

ELSE

no failure and the material point stiffness should

not be deleted.

STATE(12) = 0

Implementation in EUROPLEXUS

* update state variables before exit

STATE(1) = SIGMAH

STATE(2) = PHI

STATE(3) = P

STATE(4) = PHITRIAL

STATE(5) = F

STATE(6) = R

STATE(7) = DDLAMBDA

STATE(8) = DLAMBDA

STATE(9) = NRITER

STATE(10) = DLAMBDA / DT

STATE(11) = D

*

STATE(13) = T

STATE(14) = WE

STATE(15) = CSON

*

STATE(16) = S1

STATE(17) = S2

STATE(18) = S3

*

* decide if material point stiffness is zero, i.e. failed

*--------------------------------------------------------------

990 IF (D >= DC) THEN

*

* the current gp was already failed or is failing now :

* state(1:10,12,13), the relevant components

* of stressnew and cson are set to 0.

* the total strains (strain) are not set to 0 so they continue to be

* updated as long as the element is not eroded, i.e. as long

* as the element has some virgin gauss points.

*

SELECT CASE (ITAU)

CASE (0)

STRESSNEW(1:3) = R0

CASE (1)

STRESSNEW(1:4) = R0

CASE (2)

STRESSNEW(1:5) = R0

CASE (3)

STRESSNEW(1:6) = R0

END SELECT

STATE(1:10) = R0

* state(11) (damage d) keeps last value it had before failure

STATE(12) = R0

* state(13) (absolute temperature t) keeps last value before failure

* state(14) (damage accumulation we) keeps last value before failure

CSON = R0

STATE(15) = CSON

ELSE

*

* no failure

*

STATE(12) = R1

*

66

12. Update the actual total strain ε33,n+1 and incremental strain ∆ε33,n+1 through the thickness accordingto Eq. (69) and return this to the element routine in EUROPLEXUS

Goverening equations

IF plane stress THEN use Eq.(69) to compute ac-tual strain trough thickness

∆ε33,n+1 =1 − 2ν

E(∆σ11,n+1 + ∆σ22,n+1)

− (∆ε11,n+1 + ∆ε22,n+1)

ε33,n+1 = ε33,n + ∆ε33,n+1

Implementation in EUROPLEXUS

* strain increment(s) and total strain(s) corresponding to

* the imposed zero stress condition(s)

*

SELECT CASE (IPLANC)

CASE (0) ! NO PLANE STRESS CONDITION

CASE (1) ! PLANE STRESS CONDITION (STRESS(3)=0)

* store provisional strain increment

AUX3 = STRAININC(3)

* compute and update the actual strain increment

FACTOR = T2 / YOUNG

STRAININC(3) = FACTOR*(STRESSNEW(1) - STRESSOLD(1)

> + STRESSNEW(2) - STRESSOLD(2))

> - STRAININC(1) - STRAININC(2)

* update the total strain

STRAIN(3) = STRAIN(3) - AUX3 + STRAININC(3)

CASE (2) ! UNIAXIAL STRESS CONDITION (STRESS(2)=STRESS(3)=0)

AUX2 = STRAININC(2)

AUX3 = STRAININC(3)

FACTOR = T2 / YOUNG

STRAININC(2) = 0.5D0*(FACTOR*(STRESSNEW(1) - STRESSOLD(1))

> - STRAININC(1))

STRAININC(3) = STRAININC(2)

STRAIN(2) = STRAIN(2) - AUX2 + STRAININC(2)

STRAIN(3) = STRAIN(3) - AUX3 + STRAININC(3)

END SELECT

ENDIF

*

* erosion (optional)

*

IF (L_EROSION) THEN

IF (D >= DC) FAIL_FLAG = .TRUE.

ENDIF

*

END SUBROUTINE VPJC

In EPX, this is done for each material integration point in the FE assembly. The stresses are updatedin the material interface using a fully vectorized version of the forward Euler integration algorithm anda two-state architecture where the initial values at tn are stored in the old arrays and the new values attn+1 must be updated and stored in the new arrays returned to the global FE analysis. The VPJC ma-terial routine is valid for 1D, 2D and 3D stress states, i.e., for bar, shell, solid, axisymmetric, plane strainand plane stress elements. The stresses for 3D elements are stored similar to that of symmetric tensorsas σ = [σ11, σ22, σ33, σ12, σ23, σ31]T , and plane stress, axisymmetric and plane strain elements are stored asσ = [σ11, σ22, σ33, σ12]T . The deformation gradient and strains are stored similarly to the stresses, however,one should be aware of that the shear strain is stored as engineering shear strains, e.g. γ12 = 2ε12.

Appendix F. Material routine in EUROPLEXUS

The constitutive model of elastic-thermoviscoplasticity and ductile failure is implemented in EUROPLEXUS(EPX) [7] using a fully vectorized version of the forward Euler integration algorithm and a special case of thebackward Euler integration algorithm as presented in this report. The FORTRAN implementation of thenumerical scheme in Appendix E is presented in this section. The model is applicable for 1D, 2D and 3Dstress states, i.e., for 1D, 2D and 3D stress states, i.e., for bar, shell, solid, axisymmetric, plane strain andplane stress elements. The model is coupled with an element deletion option available in EPX by introducinga state variable controlling the element erosion, i.e., as the damage variable reaches its critical value DC thestiffness in the material point is removed. An element will be removed from the mesh when a user-definedfraction of the material points in the element are deleted. This is specified in the input file for EPX usingthe EROS directive where the user should specify a (proportional) number between 0 and 1. The defaultvalue is 1 and indicates that an element is eroded when all its Gauss points have failed, while the value 0indicate that an element is eroded when any one of its Gauss points fails.

The procedure described in Section 5.1 and Appendix E is formulated in terms of the material subroutineVPJC, which communicates with the element routine (or even a higher-order routine) in EPX. During an

67

FE analysis the material subroutine VPJC is called by the chosen element routine for each material pointat each increment. When the material subroutine is called, it is provided with the state at the start ofthe increment (stress and solution-dependent state variables at tn). Based on the geometrical calculationsperformed by the element routine it is also provided with the strain increment ∆ε and total strains ε at thebeginning and the end of the increment. Thus, the VPJC uses a two-state architecture for the stress tensorσ where the initial values at tn are stored in the old arrays and the new values at tn+1 must be updated andstored in the new arrays returned to the global analysis in EPX. The stresses for 3D elements are storedsimilar to that of symmetric tensors as σ = [σ11, σ22, σ33, σ12, σ23, σ31]T , while plane stress, plane strainand axisymmetric elements are stored as σ = [σ11, σ22, σ33, σ12]T . The deformation gradient and strains arestored similarly to the stresses, however, one should be aware of that the shear strain is stored as tensorshear strains, e.g. ε12 = 1

2γ12. That is, ε = [ε11, ε22, ε33, 2ε12, 2ε23, 2ε31]T and ε = [ε11, ε22, ε33, 2ε12]T inthe 3D and 2D cases, respectively.

The coding of material routines in EUROPLEXUS follow FORTRAN 90 conventions. In addition to thevectorized interface, it is essential that all variables are defined and initialized properly using a routine thatreads the material constants from the input data set. Moreover, a conversion routine is used and called bythe element routine just before and after the material routine, to make the arrangement of stress and strainin VPJC compatible with the one used in the element routine. Thus, the conversation routine ensures thatall tensors have the same number of components and the same organization for a given element type.

The FORTRAN source of the material routine for EPX is listed below.

C M_MATERIAL_VPJC TOUS GVALSAMOS 16/06/06 20:09:13 #3077MODULE M_MATERIAL_VPJC

** material of type vpjc** the vpjc subroutine can be used for all stress states, by providing* appropriate values of itau and iplanc*

USE M_MATERIALS_ARRAY*

IMPLICIT NONESAVE

*PRIVATEPUBLIC ::> MAVPJC, ! PROPERTIES READING ROUTINE> VPJC ! MATERIAL ROUTINE

*CONTAINS

*===============================================================================SUBROUTINE MAVPJC (INUMLDC)

*------------------------------------------------------------------* vegard aune, folco casadei 11-14* material "vpjc" (visco-plastic johnson-cook)*------------------------------------------------------------------* mat : index of material read in the input file** organization of newmat%matval: (material properties)* ----------------------------------------------------* matval(1) - rho - material density* matval(2) - young - elastic modulus* matval(3) - nu - possion ratio* matval(4) - elas - initial yield stress* matval(5) - tol - tolerance newton iteration* matval(6) - mxit - maximum number of iterations* matval(7) - qr1 - asymptote first term voce hardening* matval(8) - cr1 - hardening parameter first term voce* matval(9) - qr2 - asymptote 2nd term voce hardening* matval(10)- cr2 - hardening parameter 2nd term voce* matval(11)- pdot - reference strain-rate* matval(12)- c - hardening parameter visco term* matval(13)- tq - taylor-quinney coefficient* matval(14)- cp - specific heat capacity solid material* matval(15)- tr - room temperature

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* matval(16)- tm - melting temperature* matval(17)- m - hardening parameter temperature term* matval(18)- dc - upper limit of d when softening* matval(19)- wc - cockcroft-latham failure criterion* matval(20)- solu - 1=cutting plane (default) explicit* requires N-R iterations,* 2=radial return (implicit) requires N-R iterations*

INCLUDE ’NONE.INC’*

INCLUDE ’CARMA.INC’INCLUDE ’CONSTA.INC’INCLUDE ’CUNIT.INC’INCLUDE ’REDCOM.INC’

*INTEGER, INTENT(IN) :: INUMLDC

** local variables*

INTEGER, PARAMETER :: IFIX = 4*

INTEGER, PARAMETER :: NMOT = 20*

CHARACTER*4 :: MOT(NMOT)INTEGER :: KOPT(NMOT), I

*DATA MOT /’RO ’,’YOUN’,’NU ’,’ELAS’,’TOL ’,> ’MXIT’,’QR1 ’,’CR1 ’,’QR2 ’,’CR2 ’,> ’PDOT’,’C ’,’TQ ’,’CP ’,’TR ’,> ’TM ’,’M ’,’DC ’,’WC ’,’SOLU’/

** the syntax of this material is particularly simple, because* it contains only a sequence of keywords, each followed by a value.* the first 4 keywords are mandatory, the others are optional** create new material structure newmat of the needed length*

CALL CREATE_MATERIAL (NMOT, 0, 0)NEWMAT%NAME = ’VPJC : VISCO-PLASTIC JOHNSON-COOK’NEWMAT%TYPE = 120 ! TYPE OF THE VPJC MATERIALNEWMAT%NUMLDC = INUMLDCNEWMAT%LGECR = LGECR(NEWMAT%TYPE)

*CALL LIRVAL (NMOT, MOT, NEWMAT%MATVAL, KOPT)

** input data complete ? (all keywords are mandatory except tol, mxit, tr, solu)*

DO I = 1, NMOTSELECT CASE (I)CASE (5, 6, 15, 20) ! TOL, MXIT, TR, SOLU ARE OPTIONALCASE DEFAULT ! ALL OTHERS ARE MANDATORYIF (KOPT(I) == 0) GO TO 9

END SELECTEND DO

** set default values in the optional properties which have not been read*

IF (KOPT(5) == 0) NEWMAT%MATVAL(5) = 1.E-4 ! TOLIF (KOPT(6) == 0) NEWMAT%MATVAL(6) = 20 ! MXITIF (KOPT(15) == 0) NEWMAT%MATVAL(15) = 293.D0 ! TRIF (KOPT(20) == 0) NEWMAT%MATVAL(20) = 1.D0 ! SOLU

** printouts*

WRITE(TAPOUT,1000) NEWMAT%NUMLDC*

WRITE(TAPOUT,1001) NEWMAT%MATVAL(1)WRITE(TAPOUT,1002) NEWMAT%MATVAL(2)WRITE(TAPOUT,1003) NEWMAT%MATVAL(3)WRITE(TAPOUT,1004) NEWMAT%MATVAL(4)WRITE(TAPOUT,1005) NEWMAT%MATVAL(5)WRITE(TAPOUT,1006) NINT (NEWMAT%MATVAL(6))WRITE(TAPOUT,1007) NEWMAT%MATVAL(7)WRITE(TAPOUT,1008) NEWMAT%MATVAL(8)WRITE(TAPOUT,1009) NEWMAT%MATVAL(9)WRITE(TAPOUT,1010) NEWMAT%MATVAL(10)WRITE(TAPOUT,1011) NEWMAT%MATVAL(11)

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WRITE(TAPOUT,1012) NEWMAT%MATVAL(12)WRITE(TAPOUT,1013) NEWMAT%MATVAL(13)WRITE(TAPOUT,1014) NEWMAT%MATVAL(14)WRITE(TAPOUT,1015) NEWMAT%MATVAL(15)WRITE(TAPOUT,1016) NEWMAT%MATVAL(16)WRITE(TAPOUT,1017) NEWMAT%MATVAL(17)WRITE(TAPOUT,1018) NEWMAT%MATVAL(18)WRITE(TAPOUT,1019) NEWMAT%MATVAL(19)WRITE(TAPOUT,1020) NINT (NEWMAT%MATVAL(20))

*RETURN

** errors*

9 CALL ERRMSS (’MAVPJC’,’DIRECTIVE "VPJC" INCOMPLETE’)STOP ’MAVPJC’

*1000 FORMAT(///,10X,’MATERIAL NUMBER ’,I5,’ ( VPJC )’,/)1001 FORMAT(30X,’DENSITY’,T65,’= ’,1PE12.5)1002 FORMAT(30X,’ELASTIC MODULUS’,T65,’= ’,1PE12.5)1003 FORMAT(30X,’POISSON RATIO’,T65,’= ’,1PE12.5)1004 FORMAT(30X,’INITIAL YIELD STRESS’,T65,’= ’,1PE12.5)1005 FORMAT(30X,’TOLERANCE FOR N-R ITERATIONS’,T65,’= ’,1PE12.5)1006 FORMAT(30X,’MAX NUMBER OF N-R ITERATIONS’,T65,’= ’,I6)1007 FORMAT(30X,’ASYMPTOTE FIRST TERM VOCE HARDENING’,T65,’= ’,1PE12.5)1008 FORMAT(30X,’HARDENING PARAMETER FIRST TERM VOCE’,T65,’= ’,1PE12.5)1009 FORMAT(30X,’ASYMPTOTE 2ND TERM VOCE HARDENING’,T65,’= ’,1PE12.5)1010 FORMAT(30X,’HARDENING PARAMETER 2ND TERM VOCE’,T65,’= ’,1PE12.5)1011 FORMAT(30X,’REFERENCE STRAIN RATE’,T65,’= ’,1PE12.5)1012 FORMAT(30X,’HARDENING PARAMETER VISCOUS TERM’,T65,’= ’,1PE12.5)1013 FORMAT(30X,’TAYLOR-QUINNEY COEFFICIENT’,T65,’= ’,1PE12.5)1014 FORMAT(30X,’SPECIFIC HEAT CAPACITY OF THE SOLID’,T65,’= ’,1PE12.5)1015 FORMAT(30X,’ROOM TEMPERATURE’,T65,’= ’,1PE12.5)1016 FORMAT(30X,’MELTING TEMPERATURE’,T65,’= ’,1PE12.5)1017 FORMAT(30X,’HARDENING PARAM. TEMPERATURE TERM’,T65,’= ’,1PE12.5)1018 FORMAT(30X,’UPPER LIMIT OF D WHEN SOFTENING’,T65,’= ’,1PE12.5)1019 FORMAT(30X,’COCKCROFT-LATHAM FAILURE CRITERION’,T65,’= ’,1PE12.5)1020 FORMAT(30X,’SOLUTION (1=CUT PLANE, 2=RAD.RET.)’,T65,’= ’,I6)*

END SUBROUTINE MAVPJC*-------------------------------------------------------------------------------

SUBROUTINE VPJC (ITAU, IPLANC, PROPS, STRAININC, STRESSOLD,> STRAIN, STATE, STRESSNEW, CSON, FAIL_FLAG)

** vegard aune, folco casadei 12-14** added optional radial return solution strategy (solu == 2) 02-16**erosion

USE M_FAILED_ELEMENTS ! for l_erosion*

USE M_PILOTUSE M_KAAPI_DATA ! FOR ELEMENT PRINTOUTS

*IMPLICIT NONE

** args

INTEGER, INTENT(IN) :: ITAU, IPLANCLOGICAL, INTENT(INOUT) :: FAIL_FLAGREAL(8), INTENT(IN) :: PROPS(*), STRESSOLD(*)REAL(8), INTENT(OUT) :: STRESSNEW(*), CSONREAL(8), INTENT(INOUT) :: STRAININC(*), STRAIN(*), STATE(*)

** locals

REAL(8) :: STRESS(6), RNV(6)*

REAL(8) :: DT, RHO, YOUNG, POISS, R3GREAL(8) :: C1, C2, C3, C4, C5, C6, C7, C8, C9REAL(8) :: SY, HR, RP, R, QR1, CR1, QR2, CR2REAL(8) :: TOL, PDOT, C, TQ, CP, TR, TM, MREAL(8) :: PN, SIGMAY, DENOMV, DENOMDV, SIGMAHREAL(8) :: P, DENOM, DDLAMBDA, DLAMBDA, RESNOR, FACTORREAL(8) :: PHITRIAL, PHI, F, D, T, DC, WC, AUX, WE, AUX3,> S1, S2, S3, A1, A2, A3, A4, DENOMCP, AUX2, T1, T2, T3,> PHIEFFREAL(8), PARAMETER :: R0 = 0.D0, R1 = 1.D0, R2 = 2.D0,> R3 = 3.D0, R4 = 4.D0, R5 = 0.5D0,

70

> R27=27.D0* defining the number pi

REAL(8), PARAMETER :: PI = 3.1415926535D0*

INTEGER :: MXITER, NRITER, SOLULOGICAL :: TEMP_SOFT ! IS TEMPERATURE SOFTENING ACTIVATED (IS M /= 0?)REAL(8) :: TSTAR, TSOFT

** input parameters to generalize the present material routine:* itau : number of shear stress components (stress(4),stress(5),stress(6))* 0 = no shear stress is present* 1 = only tau_xy (stress(4)) is present* 3 = all three shear stresses (stress(4:6)) are present* iplanc : 0 = no plane stress condition* 1 = plane stress condition (sigma_z=stress(3)=0)* 2 = uniaxial stress condition (sigma_y=stress(2)=0 and* sigma_z=stress(3)=0)** organization of stress and strain arrays* ----------------------------------------* stress(1)- xx : normal components* stress(2)- yy* stress(3)- zz* stress(4)- xy : shear components* stress(5)- yz* stress(6)- xz** attention: the shear components of strain and straininc arrays are the* "engineering" values (gamma), i.e. twice the "tensor" values (epsilon)!** list of material properties sent to the subroutine* --------------------------------------------------* props(1) - rho - material density* props(2) - young - elastic modulus* props(3) - poiss - possion ratio* props(4) - sy - initial yield stress* props(5) - tol - tolerance newton iteration* props(6) - mxiter - maximum number of iterations* props(7) - qr1 - asymptote first term voce hardening* props(8) - cr1 - hardening parameter first term voce* props(9) - qr2 - asymptote 2nd term voce hardening* props(10)- cr2 - hardening parameter 2nd term voce* props(11)- pdot - reference strain-rate* props(12)- c - hardening parameter visco term* props(13)- tq - taylor-quinney coefficient* props(14)- cp - specific heat capacity solid material* props(15)- tr - room temperature* props(16)- tm - melting temperature* props(17)- m - hardening parameter temperature term* props(18)- dc - upper limit of d when softening* props(19)- wc - cockcroft-latham failure criterion* props(20)- solu - 1=cutting plane (default) requires N-R iterations,* 2=radial return (direct, no iterations)* (not available in plane or uniaxial stress!)** list of state variables (old and new)* ----------------------------------------------------------------* description name intent* ----------------------------------------------------------------* state(1) - hydrostatic pressure (1/3 sigma_kk) sigmah new* state(2) - von mises equivalent stress phi new* state(3) - equivalent plastic strain p old+new* state(4) - elastic trial equivalent stress phitrial new* state(5) - yield function (should be near zero) f new* state(6) - total hardening of the material r old+new* state(7) - change of dlambda over the step ddlambda new* state(8) - incremental plastic multiplier dlambda new* state(9) - n of iterations for convergence nriter new* state(10)- plastic strain rate dlambda / dt new* state(11)- damage d old+new* state(12)- failure: 1 = virgin g.p., 0 = failed g.p. fa new* state(13)- absolute temperature t old+new* state(14)- cockcroft-latham damage accumulation we old+new* state(15)- sound speed cson new* state(16)- first principal stress s1 new* state(17)- second principal stress s2 new* state(18)- third principal stress s3 new

71

** fail_flag : output only!* in input it is always .false.* in output it is .true. if:* - the current Gauss point is failed, **and*** - erosion has been activated by the user* (i.e. l_erosion == .true.)** NOTE:* - the material routine is called also for an already failed gp, if the* element to which the gp belongs has not been eroded yet!* - however, the material routine is not called at all for an eroded element!*! Thread variable

INTEGER :: NTH*

DC = PROPS(18)D = STATE(11)T = STATE(13) ! T IS INITIALIZED TO TR IN ECROU.FF

** check previous failure of current gp

IF (D >= DC) THEN* current gauss point was previously failed, so let’s just skip it ...

GO TO 990ENDIF

** current gp was not previously failed, but it might fail now ...*

CALL PILOT_GET_THREAD_NUM (NTH)! End thread variable*

CALL ELEM_DT1 (IEL(NTH), DT)*

RHO = PROPS(1)YOUNG = PROPS(2)POISS = PROPS(3)SY = PROPS(4)TOL = PROPS(5)MXITER = NINT (PROPS(6))QR1 = PROPS(7)CR1 = PROPS(8)QR2 = PROPS(9)CR2 = PROPS(10)PDOT = PROPS(11)C = PROPS(12)TQ = PROPS(13)CP = PROPS(14)TR = PROPS(15)TM = PROPS(16)M = PROPS(17)WC = PROPS(19)SOLU = NINT (PROPS(20))

*IF (SOLU == 2) THEN

* radial return is not available in either plane stress or uniaxial stressIF (IPLANC > 0) THENCALL ERRMSS (’VPJC’, ’SOLU 2 N/A IN PLANE/UNIAXIAL STRESS’)STOP ’VPJC - SOLU 2 N/A IN PLANE/UNIAXIAL STRESS’

ENDIFENDIF

*TEMP_SOFT = (M /= 0.D0) ! TEMPERATURE SOFTENING IS ACTIVATED ONLY

! IF M /= 0 (EXPONENTIALS ARE EXPENSIVE ...)** sound speed

T1 = R1 + POISST2 = R1 - R2*POISST3 = R1 - POISS

*SELECT CASE (IPLANC)CASE (0) ! NO PLANE STRESS CONDITIONCSON = SQRT (YOUNG*T3 / (RHO*T1*T2))

CASE (1) ! PLANE STRESS CONDITION (SIG(3)=0)CSON = SQRT (YOUNG / (RHO*T1*T3))

CASE (2) ! UNIAXIAL STRESS CONDITION (SIG(2)=SIG(3)=0)CSON = SQRT (YOUNG / RHO)

CASE DEFAULT ! WE CHECK THIS ONLY IN THE FIRST SELECT CASE ON IPLANC

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CALL ERRMSS (’VPJC’, ’WRONG IPLANC’)STOP ’VPJC - WRONG IPLANC’

END SELECT**--------------------------------------------------------------* state update procedure for the von mises elasto-* thermoviscoplastic material model with non-linear isotropic* hardening goverened by the modified johnson-cook model:* explicit elastic predictor/return mapping algorithm.*--------------------------------------------------------------** elastic predictor: compute elastic trial stress state*--------------------------------------------------------------** compute the elasticity tensor

SELECT CASE (IPLANC)CASE (0) ! NO PLANE STRESS CONDITIONC1 = YOUNG*T3 / (T1*T2)C4 = POISS*C1 / T3

CASE (1) ! PLANE STRESS CONDITION (SIG(3)=0)C1 = YOUNG / (T1*T3)C4 = POISS*C1

CASE (2) ! UNIAXIAL STRESS CONDITION (SIG(2)=SIG(3)=0)C1 = YOUNGC4 = R0 ! THESE VALUES HAVE TO BE CHECKED

END SELECT*

C2 = C1C3 = C1C5 = C4C6 = C4C7 = R5*YOUNG / T1C8 = C7C9 = C7R3G = R3*C7

** compute the elastic trial stress and the elastic trial equivalent stress*

STRESS(1) = STRESSOLD(1) + C1*STRAININC(1) + C4*STRAININC(2)> + C5*STRAININC(3)

*SELECT CASE (IPLANC) ! TREAT NORMAL COMPONENTS FIRSTCASE (0) ! NO PLANE STRESS CONDITIONSTRESS(2) = STRESSOLD(2) + C4*STRAININC(1) + C2*STRAININC(2)

> + C6*STRAININC(3)STRESS(3) = STRESSOLD(3) + C5*STRAININC(1) + C6*STRAININC(2)

> + C3*STRAININC(3)PHITRIAL = STRESS(1)*STRESS(1) + STRESS(2)*STRESS(2)

> + STRESS(3)*STRESS(3) - STRESS(1)*STRESS(2)> - STRESS(2)*STRESS(3) - STRESS(3)*STRESS(1)CASE (1) ! PLANE STRESS CONDITION (STRESS(3)=0)STRESS(2) = STRESSOLD(2) + C4*STRAININC(1) + C2*STRAININC(2)

> + C6*STRAININC(3)STRESS(3) = R0PHITRIAL = STRESS(1)*STRESS(1) + STRESS(2)*STRESS(2)

> - STRESS(1)*STRESS(2)CASE (2) ! UNIAXIAL STRESS CONDITION (STRESS(2)=STRESS(3)=0)STRESS(2) = R0STRESS(3) = R0PHITRIAL = STRESS(1)*STRESS(1)

END SELECT*

SELECT CASE (ITAU) ! ADD SHEAR COMPONENTS IF PRESENTCASE (0) ! NO SHEAR STRESS IS PRESENTCASE (1) ! ONLY TAU_XY (STRESS(4)) IS PRESENTSTRESS(4) = STRESSOLD(4) + C7*STRAININC(4) ! THE GAMMAS ARE USED!PHITRIAL = PHITRIAL + R3*(STRESS(4)*STRESS(4))

CASE (3) ! ALL THREE SHEAR STRESSES (STRESS(4:6)) ARE PRESENTSTRESS(4) = STRESSOLD(4) + C7*STRAININC(4) ! SHEAR STRAIN AND STRAININCSTRESS(5) = STRESSOLD(5) + C8*STRAININC(5) ! ARE THE GAMMAS, NOTSTRESS(6) = STRESSOLD(6) + C9*STRAININC(6) ! THE EPSILONSPHITRIAL = PHITRIAL + R3*(STRESS(4)*STRESS(4)

> + STRESS(5)*STRESS(5)> + STRESS(6)*STRESS(6))CASE DEFAULT ! WE CHECK THIS ONLY IN THE FIRST SELECT CASE ON ITAUCALL ERRMSS (’VPJC’, ’WRONG ITAU’)STOP ’VPJC - WRONG ITAU’

73

END SELECT*

PHITRIAL = SQRT (PHITRIAL)** check for plastic admissibility*--------------------------------------------------------------

PN = STATE(3)RP = STATE(6)SIGMAY = SY + RP

** import old value damage variables

WE = STATE(14)*

IF (TEMP_SOFT) THENTSTAR = (T-TR) / (TM-TR)TSOFT = R1 - TSTAR**MF = PHITRIAL - SIGMAY*TSOFT

ELSE ! M = 0 --> THE CODE ASSUMES TSTAR=0, SO THAT TSOFT=1F = PHITRIAL - SIGMAY

ENDIFDLAMBDA = R0DDLAMBDA = R0IF (PHITRIAL == R0) THENPHI = R1

ELSEPHI = PHITRIAL

ENDIF*

IF (SOLU == 2) GO TO 200**=======================================================================* solu == 1 : cutting plane solution strategy (default)*=======================================================================*

IF (F > R0) THEN** plastic step: apply return mapping* the cutting plane method uses values from the previous* iteration to update the plastic parameter, dlambda.* thus, all subsequent values (until dlambda) are related to* the previous iteration, "i".*--------------------------------------------------------------

P = PNLOOP_NR : DO NRITER = 1, MXITER

** compute the respective gradients of the von mises yield surface

RNV(2:6) = R0 ! INITIALIZE TO ZERO "OPTIONAL" COMPONENTSSELECT CASE (IPLANC) ! TREAT NORMAL COMPONENTS FIRSTCASE (0) ! NO PLANE STRESS CONDITIONRNV(1) = (STRESS(1) - R5*(STRESS(2) + STRESS(3))) / PHIRNV(2) = (STRESS(2) - R5*(STRESS(3) + STRESS(1))) / PHIRNV(3) = (STRESS(3) - R5*(STRESS(1) + STRESS(2))) / PHI

CASE (1) ! PLANE STRESS CONDITION (STRESS(3)=0)RNV(1) = (STRESS(1) - R5*STRESS(2)) / PHIRNV(2) = (STRESS(2) - R5*STRESS(1)) / PHI

CASE (2) ! UNIAXIAL STRESS CONDITION (STRESS(2)=STRESS(3)=0)RNV(1) = STRESS(1) / PHI ! TO BE CHECKED ...

END SELECT*

SELECT CASE (ITAU) ! TREAT SHEAR COMPONENTSCASE (0) ! NO SHEAR STRESS IS PRESENTCASE (1) ! ONLY TAU_XY (STRESS(4)) IS PRESENTRNV(4) = R3*STRESS(4) / PHI

CASE (3) ! ALL THREE SHEAR STRESSES (STRESS(4:6)) ARE PRESENTRNV(4) = R3*STRESS(4) / PHIRNV(5) = R3*STRESS(5) / PHIRNV(6) = R3*STRESS(6) / PHI

END SELECT** compute hardening modulus

HR = QR1*CR1*EXP(-CR1*P) + QR2*CR2*EXP(-CR2*P)** compute denominator in the residual derivative of dlambda

SELECT CASE (IPLANC) ! TREAT NORMAL COMPONENTS FIRSTCASE (0) ! NO PLANE STRESS CONDITIONDENOMCP = C1*(RNV(1)*RNV(1) + RNV(2)*RNV(2) + RNV(3)*RNV(3))

> + R2*C4*(RNV(1)*RNV(2) + RNV(2)*RNV(3) + RNV(3)*RNV(1))

74

CASE (1) ! PLANE STRESS CONDITION (RNV(3)=0)DENOMCP = C1*(RNV(1)*RNV(1) + RNV(2)*RNV(2))

> + R2*C4*(RNV(1)*RNV(2))CASE (2) ! UNIAXIAL STRESS CONDITION (RNV(2)=RNV(3)=0)DENOMCP = C1*(RNV(1)*RNV(1)) ! TO BE CHECKED ...

END SELECT*

SELECT CASE (ITAU) ! TREAT SHEAR COMPONENTSCASE (0) ! NO SHEAR STRESS IS PRESENTCASE (1) ! ONLY TAU_XY (RNV(4)) IS PRESENTDENOMCP = DENOMCP + R4*C7*(RNV(4)*RNV(4))

CASE (3) ! ALL THREE SHEAR STRESSES (RNV(4:6)) ARE PRESENTDENOMCP = DENOMCP + R4*C7*(RNV(4)*RNV(4) + RNV(5)*RNV(5)

> + RNV(6)*RNV(6))END SELECT

*DENOMV = (R1 + DLAMBDA / (PDOT*DT))**CDENOMDV = (C / (PDOT*DT))*(R1 + DLAMBDA / (PDOT*DT))**(C - R1)IF (TEMP_SOFT) THENTSTAR = (T - TR) / (TM - TR)TSOFT = R1 - TSTAR**MDENOM = DENOMCP + HR*DENOMV*TSOFT + SIGMAY*DENOMDV*TSOFT

ELSE ! M = 0 --> THE CODE ASSUMES TSTAR=0, SO THAT TSOFT=1DENOM = DENOMCP + HR*DENOMV + SIGMAY*DENOMDV

ENDIF** compute residual increment ddlambda and update variable dlambda

DDLAMBDA = F / DENOMDLAMBDA = DLAMBDA + DDLAMBDA

** update internal variables dependent on dlambda

P = P + DDLAMBDAR = RP + HR*DLAMBDASIGMAY = SY + R

** update stress components using the elastic (hooke’s) law

SELECT CASE (IPLANC) ! TREAT NORMAL COMPONENTS FIRSTCASE (0) ! NO PLANE STRESS CONDITIONSTRESS(1) = STRESS(1) -

> DDLAMBDA*(C1*RNV(1) + C4*RNV(2) + C5*RNV(3))STRESS(2) = STRESS(2) -

> DDLAMBDA*(C4*RNV(1) + C2*RNV(2) + C6*RNV(3))STRESS(3) = STRESS(3) -

> DDLAMBDA*(C5*RNV(1) + C6*RNV(2) + C3*RNV(3))CASE (1) ! PLANE STRESS CONDITION (RNV(3)=0)STRESS(1) = STRESS(1) -

> DDLAMBDA*(C1*RNV(1) + C4*RNV(2))STRESS(2) = STRESS(2) -

> DDLAMBDA*(C4*RNV(1) + C2*RNV(2))CASE (2) ! UNIAXIAL STRESS CONDITION (RNV(2)=RNV(3)=0)STRESS(1) = STRESS(1) - DDLAMBDA*C1*RNV(1) ! TO BE CHECKED ...

END SELECT*

SELECT CASE (ITAU) ! TREAT SHEAR COMPONENTSCASE (0) ! NO SHEAR STRESS IS PRESENTCASE (1) ! ONLY TAU_XY (RNV(4)) IS PRESENTSTRESS(4) = STRESS(4) - DDLAMBDA*C7*RNV(4)

CASE (3) ! ALL THREE SHEAR STRESSES (RNV(4:6)) ARE PRESENTSTRESS(4) = STRESS(4) - DDLAMBDA*C7*RNV(4)STRESS(5) = STRESS(5) - DDLAMBDA*C8*RNV(5)STRESS(6) = STRESS(6) - DDLAMBDA*C9*RNV(6)

END SELECT** compute updated equivalent stress

SELECT CASE (IPLANC) ! TREAT NORMAL COMPONENTS FIRSTCASE (0) ! NO PLANE STRESS CONDITIONPHI = STRESS(1)*STRESS(1) + STRESS(2)*STRESS(2)

> + STRESS(3)*STRESS(3) - STRESS(1)*STRESS(2)> - STRESS(2)*STRESS(3) - STRESS(3)*STRESS(1)

CASE (1) ! PLANE STRESS CONDITION (STRESS(3)=0)PHI = STRESS(1)*STRESS(1) + STRESS(2)*STRESS(2)

> - STRESS(1)*STRESS(2)CASE (2) ! UNIAXIAL STRESS CONDITION (STRESS(2)=STRESS(3)=0)PHI = STRESS(1)*STRESS(1)

END SELECT*

SELECT CASE (ITAU) ! ADD SHEAR COMPONENTS IF PRESENT

75

CASE (0) ! NO SHEAR STRESS IS PRESENTCASE (1) ! ONLY TAU_XY (STRESS(4)) IS PRESENTPHI = PHI + R3*(STRESS(4)*STRESS(4))

CASE (3) ! ALL THREE SHEAR STRESSES (STRESS(4:6)) ARE PRESENTPHI = PHI + R3*(STRESS(4)*STRESS(4)

> + STRESS(5)*STRESS(5)> + STRESS(6)*STRESS(6))

END SELECT*

PHI = SQRT (PHI)** compute yield function and new residual

IF (TEMP_SOFT) THENTSTAR = (T - TR) / (TM - TR) ! THIS RECALCULATION OF TSTARTSOFT = R1 - TSTAR**M ! AND TSOFT SEEMS REDUNDANT HEREF = PHI - SIGMAY*((R1 + DLAMBDA / (PDOT*DT))**C)*TSOFT

** check convergence

RESNOR = F / (SIGMAY*((R1 + DLAMBDA / (PDOT*DT))**C)*TSOFT)RESNOR = ABS (RESNOR)

ELSE ! M = 0 --> THE CODE ASSUMES TSTAR=0, SO THAT TSOFT=1F = PHI - SIGMAY*((R1 + DLAMBDA / (PDOT*DT))**C)

* check convergenceRESNOR = F / (SIGMAY*((R1 + DLAMBDA / (PDOT*DT))**C))RESNOR = ABS (RESNOR)

ENDIF*

IF (RESNOR <= TOL) THEN** update stress components based on plastic step*-----------------------------------------------

STRESSNEW(1) = STRESS(1)STRESSNEW(2:4) = R0 ! INITIALIZE TO ZERO "OPTIONAL" COMPONENTSSELECT CASE (IPLANC) ! TREAT NORMAL COMPONENTS FIRSTCASE (0) ! NO PLANE STRESS CONDITIONSTRESSNEW(2) = STRESS(2)STRESSNEW(3) = STRESS(3)

CASE (1) ! PLANE STRESS CONDITION (STRESS(3)=0)STRESSNEW(2) = STRESS(2)

CASE (2) ! UNIAXIAL STRESS CONDITION (STRESS(2)=STRESS(3)=0)END SELECT

*SELECT CASE (ITAU) ! ADD SHEAR COMPONENTS IF PRESENTCASE (0) ! NO SHEAR STRESS IS PRESENTCASE (1) ! ONLY TAU_XY (STRESS(4)) IS PRESENTSTRESSNEW(4) = STRESS(4)

CASE (3) ! ALL THREE SHEAR STRESSES (STRESS(4:6)) ARE PRESENTSTRESSNEW(4) = STRESS(4)STRESSNEW(5) = STRESS(5)STRESSNEW(6) = STRESS(6)

END SELECT** update remaining variables dependent of dlambda*------------------------------------------------* update temperature (only if temperature softening is activated)

IF (TEMP_SOFT) THENT = T + TQ*PHI*DLAMBDA / (RHO*CP)IF (T >= TM) THEN ! CHECK IF MELTING OCCURSSTATE(13) = TMD = DC ! this will cause the GP to fail (see 990)GO TO 990

ENDIFENDIF

** compute hydrostatic stress based on new stress state,

SIGMAH = STRESS(1)SELECT CASE (IPLANC)CASE (0) ! NO PLANE STRESS CONDITIONSIGMAH = SIGMAH + STRESS(2) + STRESS(3)

CASE (1) ! PLANE STRESS CONDITION (STRESS(3)=0)SIGMAH = SIGMAH + STRESS(2)

CASE (2) ! UNIAXIAL STRESS CONDITION (STRESS(2)=STRESS(3)=0)END SELECT

*SIGMAH = SIGMAH / R3

** 3d stress state and compute deviatoric stress

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S1 = STRESS(1) - SIGMAHS2 = STRESS(2) - SIGMAHS3 = STRESS(3) - SIGMAH

** finding principal stress based on the deviatoric stress and invariants

A1 = R5*(S1*S1 + S2*S2 + S3*S3)*

SELECT CASE (ITAU) ! ADD SHEAR COMPONENTS IF PRESENTCASE (0) ! NO SHEAR STRESS IS PRESENTA2 = -S1*S2*S3 ! DETERMINANT OF THE DEVIATORIC STRESS TENSOR

CASE (1) ! ONLY TAU_XY (STRESS(4)) IS PRESENTA1 = A1 + STRESS(4)*STRESS(4)A2 = S3*STRESS(4)*STRESS(4) - S1*S2*S3

CASE (3) ! ALL THREE SHEAR STRESSES (STRESS(4:6)) ARE PRESENTA1 = A1 +

> STRESS(4)*STRESS(4) + STRESS(5)*STRESS(5) +> STRESS(6)*STRESS(6)

A2 = S1*STRESS(5)*STRESS(5) +> S2*STRESS(6)*STRESS(6) +> S3*STRESS(4)*STRESS(4) -> S1*S2*S3 - R2*STRESS(4)*STRESS(5)*STRESS(6)

END SELECT*

A3 = -SQRT (R27 / A1)*A2*R5 / A1A3 = SIGN (MIN (ABS (A3), R1), A3)A4 = ACOS (A3) / R3

** compute (chronological) principal stresses

AUX = SQRT (A1 / R3)S1 = SIGMAH + R2*AUX*COS(A4)S2 = SIGMAH + R2*AUX*COS(A4-R2*PI/R3) ! only for postprocessingS3 = SIGMAH + R2*AUX*COS(A4+R2*PI/R3) ! only for postprocessing

*WE = WE + MAX (S1, R0)*DLAMBDAD = WE / WC

** update state variables before exit

STATE(1) = SIGMAHSTATE(2) = PHISTATE(3) = PSTATE(4) = PHITRIALSTATE(5) = FSTATE(6) = RSTATE(7) = DDLAMBDASTATE(8) = DLAMBDASTATE(9) = NRITERSTATE(10) = DLAMBDA / DTSTATE(11) = D

*STATE(13) = TSTATE(14) = WESTATE(15) = CSON

*STATE(16) = S1STATE(17) = S2STATE(18) = S3

*EXIT LOOP_NR

ELSE ! (RESNOR >= TOL)IF (NRITER == MXITER) THENCALL ERRMSS (’VPJC’,’MAX NO. OF N-R ITERATIONS REACHED’)STOP ’VPJC - MAX NO. OF N-R ITERATIONS REACHED’

ENDIFENDIF ! (RESNOR < TOL)

*END DO LOOP_NR

*ELSE ! (F <= R0)

** if not plasticity, elastic update

STRESSNEW(1) = STRESS(1)STRESSNEW(2:4) = R0 ! INITIALIZE TO ZERO "OPTIONAL" COMPONENTSSELECT CASE (IPLANC) ! TREAT NORMAL COMPONENTS FIRSTCASE (0) ! NO PLANE STRESS CONDITIONSTRESSNEW(2) = STRESS(2)STRESSNEW(3) = STRESS(3)

CASE (1) ! PLANE STRESS CONDITION (STRESS(3)=0)

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STRESSNEW(2) = STRESS(2)CASE (2) ! UNIAXIAL STRESS CONDITION (STRESS(2)=STRESS(3)=0)END SELECT

*SELECT CASE (ITAU) ! ADD SHEAR COMPONENTS IF PRESENTCASE (0) ! NO SHEAR STRESS IS PRESENTCASE (1) ! ONLY TAU_XY (STRESS(4)) IS PRESENTSTRESSNEW(4) = STRESS(4)

CASE (3) ! ALL THREE SHEAR STRESSES (STRESS(4:6)) ARE PRESENTSTRESSNEW(4) = STRESS(4)STRESSNEW(5) = STRESS(5)STRESSNEW(6) = STRESS(6)

END SELECT*

SIGMAH = STRESS(1)SELECT CASE (IPLANC)CASE (0) ! NO PLANE STRESS CONDITIONSIGMAH = SIGMAH + STRESS(2) + STRESS(3)

CASE (1) ! PLANE STRESS CONDITION (STRESS(3)=0)SIGMAH = SIGMAH + STRESS(2)

CASE (2) ! UNIAXIAL STRESS CONDITION (STRESS(2)=STRESS(3)=0)END SELECT

*SIGMAH = SIGMAH / R3

** 3d stress state and compute deviatoric stress

S1 = STRESS(1) - SIGMAHS2 = STRESS(2) - SIGMAHS3 = STRESS(3) - SIGMAH

** finding principal stress based on the deviatoric stress and invariants

A1 = R5*(S1*S1 + S2*S2 + S3*S3)*

SELECT CASE (ITAU) ! ADD SHEAR COMPONENTS IF PRESENTCASE (0) ! NO SHEAR STRESS IS PRESENTA2 = -S1*S2*S3 ! DETERMINANT OF THE DEVIATORIC STRESS TENSOR

CASE (1) ! ONLY TAU_XY (STRESS(4)) IS PRESENTA1 = A1 + STRESS(4)*STRESS(4)A2 = S3*STRESS(4)*STRESS(4) - S1*S2*S3

CASE (3) ! ALL THREE SHEAR STRESSES (STRESS(4:6)) ARE PRESENTA1 = A1 +

> STRESS(4)*STRESS(4) + STRESS(5)*STRESS(5) +> STRESS(6)*STRESS(6)

A2 = S1*STRESS(5)*STRESS(5) +> S2*STRESS(6)*STRESS(6) +> S3*STRESS(4)*STRESS(4) -> S1*S2*S3 - R2*STRESS(4)*STRESS(5)*STRESS(6)

END SELECT*

A3 = -SQRT (R27 / A1)*A2*R5 / A1A3 = SIGN (MIN (ABS (A3), R1), A3)A4 = ACOS (A3) / R3

** compute (chronological) principal stresses

AUX = SQRT (A1 / R3)S1 = SIGMAH + R2*AUX*COS(A4)S2 = SIGMAH + R2*AUX*COS(A4-R2*PI/R3) ! only for postprocessingS3 = SIGMAH + R2*AUX*COS(A4+R2*PI/R3) ! only for postprocessing

*STATE(1) = SIGMAH ! COMPUTE SIGMAH AND INSERT IT HERESTATE(2) = PHITRIAL ! COMPUTE PHI AND INSERT IT HERESTATE(3) = PN ! COMPUTE P AND INSERT IT HERESTATE(4) = PHITRIALSTATE(5) = FSTATE(6) = RPSTATE(7) = R0STATE(8) = R0STATE(9) = R0STATE(10) = R0STATE(11) = DSTATE(12) = R1STATE(13) = TSTATE(14) = WESTATE(15) = CSON

*STATE(16) = S1STATE(17) = S2

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STATE(18) = S3*

ENDIF ! (F > R0)*

GO TO 990*200 CONTINUE

**=======================================================================* solu == 2 : radial return solution strategy*=======================================================================*C COMPUTE THE GRADIENT OF THE VON MISES YIELD SURFACE

RNV(2:6) = R0 ! INITIALIZE TO ZERO "OPTIONAL" COMPONENTSSELECT CASE (IPLANC) ! TREAT NORMAL COMPONENTS FIRSTCASE (0) ! NO PLANE STRESS CONDITIONRNV(1) = (STRESS(1) - R5*(STRESS(2) + STRESS(3))) / PHIRNV(2) = (STRESS(2) - R5*(STRESS(3) + STRESS(1))) / PHIRNV(3) = (STRESS(3) - R5*(STRESS(1) + STRESS(2))) / PHI

CASE (1) ! PLANE STRESS CONDITION (STRESS(3)=0)* THIS CASE IS ILLEGAL FOR SOLU == 2 (SEE TEST AT BEGINNING OF ROUTINE)

RNV(1) = (STRESS(1) - R5*STRESS(2)) / PHIRNV(2) = (STRESS(2) - R5*STRESS(1)) / PHI

CASE (2) ! UNIAXIAL STRESS CONDITION (STRESS(2)=STRESS(3)=0)* THIS CASE IS ILLEGAL FOR SOLU == 2 (SEE TEST AT BEGINNING OF ROUTINE)

RNV(1) = STRESS(1) / PHI ! TO BE CHECKED ...END SELECT

*SELECT CASE (ITAU) ! TREAT SHEAR COMPONENTSCASE (0) ! NO SHEAR STRESS IS PRESENTCASE (1) ! ONLY TAU_XY (STRESS(4)) IS PRESENTRNV(4) = R3*STRESS(4) / PHI

CASE (3) ! ALL THREE SHEAR STRESSES (STRESS(4:6)) ARE PRESENTRNV(4) = R3*STRESS(4) / PHIRNV(5) = R3*STRESS(5) / PHIRNV(6) = R3*STRESS(6) / PHI

END SELECT*

IF (F > R0) THEN*C PLASTIC STEP: APPLY RADIAL RETURN MAPPING - USE NEWTON-RAPHSONC ALGORITHM TO SOLVE THE RETURN MAPPING EQUATIONC--------------------------------------------------------------

P = PNLOOP_NR2 : DO NRITER = 1, MXITER

*C COMPUTE HARDENING MODULUS

HR = QR1*CR1*EXP(-CR1*P) + QR2*CR2*EXP(-CR2*P)C COMPUTE RESIDUAL DERIVATIVE

DENOMV = (R1 + DLAMBDA/(PDOT*DT))**CDENOMDV = (C/(PDOT*DT))*(R1+DLAMBDA/(PDOT*DT))**(C-R1)IF (TEMP_SOFT) THENTSTAR = (T - TR) / (TM - TR)TSOFT = R1 - TSTAR**M

ELSE ! M = 0 --> THE CODE ASSUMES TSTAR=0, SO THAT TSOFT=1TSTAR = R0TSOFT = R1

ENDIFDENOM = R3G + HR*DENOMV*TSOFT + SIGMAY*DENOMDV*TSOFT

C COMPUTE NEWTON-RAPHSON INCREMENT AND UPDATE VARIABLE DLAMBDADDLAMBDA = F / DENOMDLAMBDA = DLAMBDA + DDLAMBDA

C COMPUTE NEW RESIDUALP = P + DDLAMBDAR = RP + HR*DLAMBDASIGMAY = SY + RDENOMV = (R1 + DLAMBDA/(PDOT*DT))**CF = PHITRIAL - R3G*DLAMBDA - SIGMAY*DENOMV*TSOFT

C CHECK CONVERGENCERESNOR = F / (SIGMAY*DENOMV*TSOFT)RESNOR = ABS (RESNOR)

*IF (RESNOR <= TOL) THEN

*C UPDATE VARIABLES DEPENDENT OF DLAMBDAC--------------------------------------------------------------C UPDATE TEMPERATURE (ONLY IF TEMPERATURE SOFTENING IS ACTIVATED)

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IF (TEMP_SOFT) THENPHIEFF = PHITRIAL - R3G*DLAMBDA ! EQUIVALENT (EFFECTIVE) STRESST = T + TQ*PHIEFF*DLAMBDA / (RHO*CP)IF (T >= TM) THEN ! CHECK IF MELTING OCCURS

STATE(13) = TMD = DC ! this will cause the GP to fail (see 990)GO TO 990

ENDIFENDIF

C COMPUTE HYDROSTATIC STRESS BASED ON TRIAL STRESS STATE,C I.E. UTILIZE THAT THE HYDROSTATIC STRESS IS THE SAMEC FOR BOTH TRIAL STATE AND FINAL STATE. ONLY THE DEVIATORICC STRESS STATE WILL CHANGE.

SIGMAH = STRESS(1)SELECT CASE (IPLANC)CASE (0) ! NO PLANE STRESS CONDITIONSIGMAH = SIGMAH + STRESS(2) + STRESS(3)

CASE (1) ! PLANE STRESS CONDITION (STRESS(3)=0)* THIS CASE IS ILLEGAL FOR SOLU == 2 (SEE TEST AT BEGINNING OF ROUTINE)

SIGMAH = SIGMAH + STRESS(2)CASE (2) ! UNIAXIAL STRESS CONDITION (STRESS(2)=STRESS(3)=0)

* THIS CASE IS ILLEGAL FOR SOLU == 2 (SEE TEST AT BEGINNING OF ROUTINE)END SELECTSIGMAH = SIGMAH / R3

C COMPUTE DEVIATORIC STRESSS1 = STRESS(1) - SIGMAHS2 = STRESS(2) - SIGMAHS3 = STRESS(3) - SIGMAH

C FINDING PRINCIPAL STRESS BASED ON THE DEVIATORIC STRESS AND INVARIANTSA1 = R5*(S1*S1 + S2*S2 + S3*S3)SELECT CASE (ITAU) ! ADD SHEAR COMPONENTS IF PRESENTCASE (0) ! NO SHEAR STRESS IS PRESENTA2 = -S1*S2*S3 ! DETERMINANT OF THE DEVIATORIC STRESS TENSOR

CASE (1) ! ONLY TAU_XY (STRESS(4)) IS PRESENTA1 = A1 + STRESS(4)*STRESS(4)A2 = S3*STRESS(4)*STRESS(4) - S1*S2*S3

CASE (3) ! ALL THREE SHEAR STRESSES (STRESS(4:6)) ARE PRESENTA1 = A1 +

> STRESS(4)*STRESS(4) + STRESS(5)*STRESS(5) +> STRESS(6)*STRESS(6)

A2 = S1*STRESS(5)*STRESS(5) +> S2*STRESS(6)*STRESS(6) +> S3*STRESS(4)*STRESS(4) -> S1*S2*S3 - R2*STRESS(4)*STRESS(5)*STRESS(6)

END SELECTA3 = -SQRT (R27 / A1)*A2*R5 / A1A3 = SIGN (MIN (ABS (A3), R1), A3)A4 = ACOS (A3) / R3

C COMPUTE (CHRONOLOGICAL) PRINCIPAL STRESSESAUX = SQRT (A1 / R3)S1 = SIGMAH + R2*AUX*COS(A4)S2 = SIGMAH + R2*AUX*COS(A4-R2*PI/R3) ! only for postprocessingS3 = SIGMAH + R2*AUX*COS(A4+R2*PI/R3) ! only for postprocessingWE = WE + MAX (S1, R0)*DLAMBDAD = WE / WC

*C UPDATE STRESS COMPONENTS USING THE ELASTIC (HOOKE’S) LAWC--------------------------------------------------------------

SELECT CASE (IPLANC) ! TREAT NORMAL COMPONENTS FIRSTCASE (0) ! NO PLANE STRESS CONDITIONSTRESSNEW(1) = STRESS(1) -

> DLAMBDA*(C1*RNV(1) + C4*RNV(2) + C5*RNV(3))STRESSNEW(2) = STRESS(2) -

> DLAMBDA*(C4*RNV(1) + C2*RNV(2) + C6*RNV(3))STRESSNEW(3) = STRESS(3) -

> DLAMBDA*(C5*RNV(1) + C6*RNV(2) + C3*RNV(3))CASE (1) ! PLANE STRESS CONDITION (RNV(3)=0)

* THIS CASE IS ILLEGAL FOR SOLU == 2 (SEE TEST AT BEGINNING OF ROUTINE)STRESSNEW(1) = STRESS(1) -

> DLAMBDA*(C1*RNV(1) + C4*RNV(2))STRESSNEW(2) = STRESS(2) -

> DLAMBDA*(C4*RNV(1) + C2*RNV(2))CASE (2) ! UNIAXIAL STRESS CONDITION (RNV(2)=RNV(3)=0)

* THIS CASE IS ILLEGAL FOR SOLU == 2 (SEE TEST AT BEGINNING OF ROUTINE)STRESSNEW(1) = STRESS(1) -

> DLAMBDA*C1*RNV(1) ! TO BE CHECKED ...END SELECT

80

*SELECT CASE (ITAU) ! TREAT SHEAR COMPONENTSCASE (0) ! NO SHEAR STRESS IS PRESENTCASE (1) ! ONLY TAU_XY (RNV(4)) IS PRESENTSTRESSNEW(4) = STRESS(4) - DLAMBDA*C7*RNV(4)

CASE (3) ! ALL THREE SHEAR STRESSES (RNV(4:6)) ARE PRESENTSTRESSNEW(4) = STRESS(4) - DLAMBDA*C7*RNV(4)STRESSNEW(5) = STRESS(5) - DLAMBDA*C8*RNV(5)STRESSNEW(6) = STRESS(6) - DLAMBDA*C9*RNV(6)

END SELECT*C COMPUTE UPDATED EQUIVALENT STRESS

SELECT CASE (IPLANC) ! TREAT NORMAL COMPONENTS FIRSTCASE (0) ! NO PLANE STRESS CONDITIONPHI = STRESSNEW(1)*STRESSNEW(1) +

> STRESSNEW(2)*STRESSNEW(2) +> STRESSNEW(3)*STRESSNEW(3) -> STRESSNEW(1)*STRESSNEW(2) -> STRESSNEW(2)*STRESSNEW(3) -> STRESSNEW(3)*STRESSNEW(1)

CASE (1) ! PLANE STRESS CONDITION (STRESS(3)=0)PHI = STRESSNEW(1)*STRESSNEW(1) +

> STRESSNEW(2)*STRESSNEW(2) -> STRESSNEW(1)*STRESSNEW(2)

CASE (2) ! UNIAXIAL STRESS CONDITION (STRESS(2)=STRESS(3)=0)PHI = STRESSNEW(1)*STRESSNEW(1)

END SELECT*

SELECT CASE (ITAU) ! ADD SHEAR COMPONENTS IF PRESENTCASE (0) ! NO SHEAR STRESS IS PRESENTCASE (1) ! ONLY TAU_XY (STRESSNEW(4)) IS PRESENTPHI = PHI + R3*(STRESSNEW(4)*STRESSNEW(4))

CASE (3) ! ALL THREE SHEAR STRESSES (STRESSNEW(4:6)) ARE PRESENTPHI = PHI + R3*(STRESSNEW(4)*STRESSNEW(4)

> + STRESSNEW(5)*STRESSNEW(5)> + STRESSNEW(6)*STRESSNEW(6))

END SELECT*

PHI = SQRT (PHI)*C UPDATE STATE VARIABLES (SDV) BEFORE EXIT

STATE(1) = SIGMAHSTATE(2) = PHISTATE(3) = PSTATE(4) = PHITRIALSTATE(5) = FSTATE(6) = RSTATE(7) = DDLAMBDASTATE(8) = DLAMBDASTATE(9) = NRITERSTATE(10) = DLAMBDA / DTSTATE(11) = D

*STATE(13) = TSTATE(14) = WESTATE(15) = CSON

*STATE(16) = S1STATE(17) = S2STATE(18) = S3

*EXIT LOOP_NR2

ELSE ! (RESNOR >= TOL)IF (NRITER == MXITER) THENCALL ERRMSS (’VPJC’,’MAX NO. OF N-R ITERATIONS REACHED’)STOP ’VPJC - MAX NO. OF N-R ITERATIONS REACHED’

ENDIFENDIF ! (RESNOR < TOL)

*END DO LOOP_NR2

*ELSE ! (F <= R0)

*C IF NOT PLASTICITY, ELASTIC UPDATE

STRESSNEW(1) = STRESS(1)STRESSNEW(2:4) = R0 ! INITIALIZE TO ZERO "OPTIONAL" COMPONENTSSELECT CASE (IPLANC) ! TREAT NORMAL COMPONENTS FIRST

81

CASE (0) ! NO PLANE STRESS CONDITIONSTRESSNEW(2) = STRESS(2)STRESSNEW(3) = STRESS(3)

CASE (1) ! PLANE STRESS CONDITION (STRESS(3)=0)* THIS CASE IS ILLEGAL FOR SOLU == 2 (SEE TEST AT BEGINNING OF ROUTINE)

STRESSNEW(2) = STRESS(2)CASE (2) ! UNIAXIAL STRESS CONDITION (STRESS(2)=STRESS(3)=0)

* THIS CASE IS ILLEGAL FOR SOLU == 2 (SEE TEST AT BEGINNING OF ROUTINE)END SELECT

*SELECT CASE (ITAU) ! ADD SHEAR COMPONENTS IF PRESENTCASE (0) ! NO SHEAR STRESS IS PRESENTCASE (1) ! ONLY TAU_XY (STRESS(4)) IS PRESENTSTRESSNEW(4) = STRESS(4)

CASE (3) ! ALL THREE SHEAR STRESSES (STRESS(4:6)) ARE PRESENTSTRESSNEW(4) = STRESS(4)STRESSNEW(5) = STRESS(5)STRESSNEW(6) = STRESS(6)

END SELECT*

SIGMAH = STRESS(1)SELECT CASE (IPLANC)CASE (0) ! NO PLANE STRESS CONDITIONSIGMAH = SIGMAH + STRESS(2) + STRESS(3)

CASE (1) ! PLANE STRESS CONDITION (STRESS(3)=0)* THIS CASE IS ILLEGAL FOR SOLU == 2 (SEE TEST AT BEGINNING OF ROUTINE)

SIGMAH = SIGMAH + STRESS(2)CASE (2) ! UNIAXIAL STRESS CONDITION (STRESS(2)=STRESS(3)=0)

* THIS CASE IS ILLEGAL FOR SOLU == 2 (SEE TEST AT BEGINNING OF ROUTINE)END SELECTSIGMAH = SIGMAH / R3

** 3d stress state and compute deviatoric stress

S1 = STRESS(1) - SIGMAHS2 = STRESS(2) - SIGMAHS3 = STRESS(3) - SIGMAH

** finding principal stress based on the deviatoric stress and invariants

A1 = R5*(S1*S1 + S2*S2 + S3*S3)*

SELECT CASE (ITAU) ! ADD SHEAR COMPONENTS IF PRESENTCASE (0) ! NO SHEAR STRESS IS PRESENTA2 = -S1*S2*S3 ! DETERMINANT OF THE DEVIATORIC STRESS TENSOR

CASE (1) ! ONLY TAU_XY (STRESS(4)) IS PRESENTA1 = A1 + STRESS(4)*STRESS(4)A2 = S3*STRESS(4)*STRESS(4) - S1*S2*S3

CASE (3) ! ALL THREE SHEAR STRESSES (STRESS(4:6)) ARE PRESENTA1 = A1 +

> STRESS(4)*STRESS(4) + STRESS(5)*STRESS(5) +> STRESS(6)*STRESS(6)

A2 = S1*STRESS(5)*STRESS(5) +> S2*STRESS(6)*STRESS(6) +> S3*STRESS(4)*STRESS(4) -> S1*S2*S3 - R2*STRESS(4)*STRESS(5)*STRESS(6)

END SELECT*

A3 = -SQRT (R27 / A1)*A2*R5 / A1A3 = SIGN (MIN (ABS (A3), R1), A3)A4 = ACOS (A3) / R3

** compute (chronological) principal stresses

AUX = SQRT (A1 / R3)S1 = SIGMAH + R2*AUX*COS(A4)S2 = SIGMAH + R2*AUX*COS(A4-R2*PI/R3) ! only for postprocessingS3 = SIGMAH + R2*AUX*COS(A4+R2*PI/R3) ! only for postprocessing

*STATE(1) = SIGMAH ! compute SIGMAH and insert it hereSTATE(2) = PHITRIAL ! compute PHI and insert it hereSTATE(3) = PN ! compute P and insert it hereSTATE(4) = PHITRIALSTATE(5) = FSTATE(6) = RPSTATE(7) = R0STATE(8) = R0STATE(9) = R0STATE(10) = R0STATE(11) = D

82

STATE(12) = R1STATE(13) = TSTATE(14) = WESTATE(15) = CSON

*STATE(16) = S1STATE(17) = S2STATE(18) = S3

*ENDIF ! (F > R0)

** decide if material point stiffness is zero, i.e. failed*--------------------------------------------------------------990 IF (D >= DC) THEN

** the current gp was already failed or is failing now :* state(1:10,12,13), the relevant components* of stressnew and cson are set to 0.* the total strains (strain) are not set to 0 so they continue to be* updated as long as the element is not eroded, i.e. as long* as the element has some virgin gauss points.*

SELECT CASE (ITAU)CASE (0)STRESSNEW(1:3) = R0

CASE (1)STRESSNEW(1:4) = R0

CASE (2)STRESSNEW(1:5) = R0

CASE (3)STRESSNEW(1:6) = R0

END SELECTSTATE(1:10) = R0

* state(11) (damage d) keeps last value it had before failureSTATE(12) = R0

* state(13) (absolute temperature t) keeps last value it had before failure* state(14) (damage accumulation we) keeps last value it had before failure

CSON = R0STATE(15) = CSON

ELSE** no failure*

STATE(12) = R1** strain increment(s) and total strain(s) corresponding to* the imposed zero stress condition(s)*

SELECT CASE (IPLANC)CASE (0) ! NO PLANE STRESS CONDITIONCASE (1) ! PLANE STRESS CONDITION (STRESS(3)=0)

* store provisional strain incrementAUX3 = STRAININC(3)

* compute and update the actual strain incrementFACTOR = T2 / YOUNGSTRAININC(3) = FACTOR*(STRESSNEW(1) - STRESSOLD(1)

> + STRESSNEW(2) - STRESSOLD(2))> - STRAININC(1) - STRAININC(2)

* update the total strainSTRAIN(3) = STRAIN(3) - AUX3 + STRAININC(3)

CASE (2) ! UNIAXIAL STRESS CONDITION (STRESS(2)=STRESS(3)=0)AUX2 = STRAININC(2)AUX3 = STRAININC(3)FACTOR = T2 / YOUNGSTRAININC(2) = 0.5D0*(FACTOR*(STRESSNEW(1) - STRESSOLD(1))

> - STRAININC(1))STRAININC(3) = STRAININC(2)STRAIN(2) = STRAIN(2) - AUX2 + STRAININC(2)STRAIN(3) = STRAIN(3) - AUX3 + STRAININC(3)

END SELECTENDIF

** erosion (optional)*

IF (L_EROSION) THENIF (D >= DC) FAIL_FLAG = .TRUE.

ENDIF

83

*END SUBROUTINE VPJC

*===============================================================================END MODULE M_MATERIAL_VPJC

84

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2

doi:10.2788/609529

ISBN 978-92-79-59746-6

LB-N

A-27982-EN

-N

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